On Well-Posedness of Nonlocal Evolution Equations

Abstract

This work studies questions of existence, uniqueness, dependence on initial data, and regularity of solutions to the Cauchy problem for nonlocal evolution equations with data in Sobolev spaces. The focus is on integrable Camassa–Holm type equations and in particular the Novikov equation and its dispersive modification. These equations apart from being interesting on their own right, also they can serve as “toy” models for the Euler equations.

1 Introduction and Results

In this work, we investigate questions of existence, uniqueness, dependence on initial data, and regularity of solutions to the initial value problem of nonlocal evolution equations. Our focus is on Camassa–Holm type equations with initial data in Sobolev spaces. We begin with four integrable equations, which Vladimir Novikov [69] derived in a unified way by examining the question of integrability for Camassa–Holm type equations of the form

$$\begin{aligned} (1-\partial _x^2)u_t = P(u, u_x, u_{xx}, u_{xxx}, \dots ), \end{aligned}$$

where P is a polynomial of u and its x-derivatives. Using as definition of integrability the existence of an infinite hierarchy of (quasi-) local higher symmetries, he produced about 20 integrable equations with quadratic nonlinearities that include the Camassa–Holm (CH) equation (see [14])

$$\begin{aligned} \partial _t u + \frac{1}{2} \partial _{x}(u^2) + (1 -\partial _{x}^{2})^{-1} \partial _{x} \left[ u^2 +\frac{1}{2}(\partial _{x}u)^2 \right] =0 \end{aligned}$$

and the Degasperis–Procesi (DP) equation (see [24])

$$\begin{aligned} \partial _t u + u\partial _{x} u + (1-\partial _x^2)^{-1} \partial _{x} \left[ \frac{3}{2} u^2\right] = 0, \end{aligned}$$

both written in their nonlocal evolutionary form. Also, he produced about 10 integrable equations with cubic nonlinearities that include the following one

$$\begin{aligned} \partial _t u + \frac{1}{3} \partial _x(u^3) + (1-\partial _x^2)^{-1}\partial _x\left[ u^3 + \frac{3}{2}u(\partial _x u)^2\right] + (1-\partial _x^2)^{-1}\left[ \frac{1}{2} (\partial _xu)^3\right] = 0, \end{aligned}$$

which is called the Novikov equation (NE) and the Fokas–Olver–Rosenau–Qiao (FORQ) equation

$$\begin{aligned} \partial _t u + \frac{1}{3} \partial _x(u^3) - \frac{1}{3} (\partial _x u)^3 + (1-\partial _x^2)^{-1}\partial _x \left[ \frac{2}{3} u^3 + u u_x^2 \right] + (1-\partial _x^2)^{-1}\partial _x\left[ \frac{1}{3} u_x^3 \right] = 0, \end{aligned}$$

which was derived earlier in [28, 70] and [72], and which is the only non-quasilinear integrable CH type equation considered here.

The CH equation arose initially in the context of hereditary symmetries studied by Fuchssteiner and Fokas [29]. However, it was written explicitly as a water wave equation by Camassa and Holm [14], who derived it from the Euler equations of hydrodynamics using asymptotic expansions. Also, they derived its peakon solutions. The existence of peakon solutions is a common phenomenon of CH, DP, NE and FORQ. On the line, these are of the form

$$\begin{aligned} \begin{array}{l} \text {CH, DP: }u(x, t)=ce^{-|x-ct|},~~ \text {NE: }u(x, t)= \sqrt{c}e^{-|x-ct|},\\ \displaystyle \text {FORQ: } u(x, t)= \sqrt{\frac{3c}{2}}e^{-|x-ct|}. \end{array} \end{aligned}$$

(1.1)

In fact, the discovery of CH by Camassa and Holm in 1993 was partly driven by the desire to find a water wave equation which has traveling wave solutions that “break” (see [76]). Recall, that the celebrated Korteweg–de Vries (KdV) equation [10, 57]

$$\begin{aligned} \partial _t u + 6 u\partial _x u + \partial ^3_x u = 0, \end{aligned}$$

which is a model of long water waves propagating in a channel, has only smooth solitons like

$$ u(x, t) = \frac{c}{2} \textrm{sech}^2 \left( \frac{\sqrt{c}}{2}(x-ct) \right) . $$

Also, CH, DP and NE have multipeakon solutions of the form

$$ u(x, t)=\sum _{j=1}^n p_j(t)e^{-|x-q_j(t)|}, $$

which will be discussed further later since they play an important role in the well-posedness theory of these equations. For more about peakon and multipeakon solutions of Camassa–Holm type equations we refer the reader to [2, 4, 5, 20, 62] and the references therein.

Next, we provide a brief survey on the well-posedness results for these equations. We begin by recalling the definition of well-posedness in Sobolev spaces on the line or the circle $H^s=H^s$ ($\mathbb {R}$ or $\mathbb {T}$) according to Hadamard, which is expressed by the following three properties:

• (i) For any initial data $u(0)\in H^s$ there exists (a lifespan) $T=T_{u(0)}>0$ and a solution $u\in C([0, T]; H^s)$ to the CH (DP, NE or FORQ) Cauchy problem.

• (ii) This solution u is unique in the Hadamard space $C([0, T]; H^s)$.

• (iii) The data-to-solution map $u(0)\mapsto u(t)$ is continuous from $H^s$ into $C([0, T]; H^s)$.

The CH and DP equations Next result states that for $s>3/2$ CH and DP are well-posed in the Sobolev spaces $H^s$ in the sense of Hadamard and the dependence of solutions on initial data is sharp, that is, it is not better than continuous.

Theorem 1.1

(CH and DP well-posedness) The Cauchy problem for the CH and DP equations is well-posed in $H^s$ for $s>3/2$. More precisely, if $s>3/2$ and $u_0 \in H^s$ then there exist $T>0$ and a unique solution $u \in C([0,T]; H^s)$ of the initial value problem for CH and DP, which depends continuously on the initial data $u_0$. Furthermore, we have the estimate

$$\begin{aligned} \Vert u(t)\Vert _{H^s} \le 2\Vert u_0\Vert _{H^s} \quad \text { for } t \ge 1/(2c_s\Vert u_0\Vert _{H^s}) \doteq T, \end{aligned}$$

(1.2)

where $c_s>0$ is a constant depending on s.

(Nonuniform Dependence) Furthermore, the data-to-solution map for these equations is not uniformly continuous from any bounded subset in $H^s$ into $C([0,T]; H^s)$.

Well-posedness of CH on the circle was proved in [44] by writing it as an ODE in a Banach space where one can prove existence and uniqueness of solutions and their continuous dependence on the initial data when these belong to a Sobolev space $H^s(\mathbb T)$ with $s>3/2$. This method follows Arnold’s approach [1] for the study of the Euler equations (see [25]). On the real line, well-posedness of CH for $s>3/2$ was proved by Li and Olver [63] using a regularization technique like in Bona and Smith [7]. The well-posedness of CH was also studied by Danchin [22], Misiolek [68] and Rodriguez-Blanco [73] using various approaches. Other works about CH can be found in [11, 12, 16,17,18,19, 23, 45, 47, 49, 50, 52, 66, 67], and [3]. For the well-posedness of DP we refer to [77] and in [34]. At this point we note that the peakon solutions u(t) defined in (1.1) belong in $H^{\frac{3}{2}^-}-H^{\frac{3}{2}}$. Therefore, these solutions are not covered by the (strong) local well posedness results mentioned in this paper. However, they are Lipchitz functions, i.e. they belong in $W^{1,\infty }$. But, the translation operator is not continuous in $W^{1,\infty }(\mathbb {R})$. So, for initial data $X\equiv H^1(\mathbb {R})\cap W^{1,\infty }(\mathbb {R})$ (containing the peakons) one can only expect to get a unique local solution in $u\in C([-T,T];H^1(\mathbb {R}))\cap L^{\infty }([-T,T];W^{1,\infty }(\mathbb {R}))$, with the data-to-solution map continuous from X to $C([-T,T];H^1(\mathbb {R}))$. For CH on the circle, this result was first established by de Lellis, Kappeler and Toplalov [23]. For CH on the line, it was proved more recently by Linares, Ponce and Sideris in [64], where they also extended the decay results obtained in [48] to the class of solutions containing peakons.

Nonuniform dependence for CH on $\mathbb {R}$ was proved in [39] by using the method of approximate solutions together with the well-posedness estimate (1.2) derived there. The periodic case was done in [40] using delicate commutator and multiplier estimates in addition to approximate solutions. For the case of DP Theorem 1.1 was proved in [34]. It is interesting that the approximate solutions used for both the CH and DP are the same. Next, we provide a brief description of the method of approximate solutions, which in the context of the Benjamin–Ono equation was first used Koch and Tzvetkov [53]. We prove that there exist two sequences of CH or DP solutions $u_n(t)$ and $v_n(t)$ in $C([0, T]; H^s(\mathbb {R}))$ satisfying the following three conditions:

• (1) $\sup _{n}{\Vert u_n(t)\Vert _{H^s}} + \sup _{n}{\Vert v_n(t)\Vert _{H^s}} \lesssim 1$,

• (2) $\lim _{n\rightarrow \infty }\Vert u_n(0) - v_n(0)\Vert _{H^s} = 0$,

• (3) $\liminf _{n}{\Vert u_n(t) - v_n(t)\Vert _{H^s}} \gtrsim \sin t$, $0\le t <T\le 1$.

We do this in the periodic case. For both the CH and DP the approximate solutions

$$\begin{aligned} u^{\omega , n}(x, t) = \omega n^{-1} + n^{-s} \cos (n x-\omega t)\quad \text { for }\omega = -1,1, \end{aligned}$$

where $n \in \mathbb {Z}^+$, satisfy conditions (1)–(3) for nonuniform dependence but they are not solutions. However, the error $E= CH(u^{\omega , n})$ or $E= DP(u^{\omega , n})$ is small in the $H^s$-norm. Then, solving the Cauchy problem with initial data $u^{\omega , n}(x, 0)$ gives actual solutions, which one proves that they satisfy the three conditions of nonuniform continuity.

Ill-posedness and norm inflation The following result shows that $s=3/2$ is critical for the well-posedness for CH and DP in Sobolev spaces.

Theorem 1.2

(Ill-posedness) For $s <3/2$, the Cauchy problem for CH and DP is not well-posed in $H^s$ in the sense of Hadamard.

For DP this result has been proved in [37]. If $1/2 \le s <3/2$, then ill-posedness is due to norm inflation. This means that there exist DP solutions who are initially arbitrarily small and eventually arbitrarily large with respect to the $H^s$ norm, in an arbitrarily short time. Since DP solutions conserve a quantity equivalent to the $L^2$-norm, there is no norm inflation in $H^0$ for these solutions. In this case, ill-posedness is caused by failure of uniqueness. For all other $s<1/2$, the situation is similar to $H^0$. For the CH we have norm inflation and thus failure of continuity for $s \in (1, \frac{3}{2})$, and failure of uniqueness for $s<1$ (see [13]). For DP norm inflation is described by the following result in [37].

Proposition 1.1

(Norm Inflation) If $s\in [\frac{1}{2}, \frac{3}{2})$, then for any $\varepsilon >0$ there is $T>0$ such that the DP Cauchy problem has a solution $u\in C([0, T); H^s)$ satisfying the following three properties:

• (1) $\textrm{Lifespan}\,T < \varepsilon$,

• (2) $\Vert u_0\Vert _{H^s} < \varepsilon$,

• (3) $\lim _{t \rightarrow T^-} \Vert u(t)\Vert _{H^s} = \infty$ (norm inflation).

The proof of Proposition 1.1 exploits the properties of appropriately constructed two-peakon solutions, which are called peakon-antipeakon solutions. We shall describe our approach for the case of the non-periodic DP equation. In this case we have that the peakon-antipeakon traveling wave

$$\begin{aligned} u(x,t) = p(t) e^{-|x+q(t)|} - p(t) e^{-|x-q(t)|} \end{aligned}$$

is a weak solution to the DP equation if the momentum $p=p(t)$ and the position $q=q(t)$ are solutions to the following system of ordinary differential equations

$$\begin{aligned} q' = p (e^{-2q} - 1) \quad \text { and } \quad p' = 2p^2 e^{-2q}. \end{aligned}$$

(1.3)

Furthermore, if for given $\varepsilon >0$ we choose $p_0$ and $q_0$ so that $p_0 \ge {1}/{\varepsilon }$ and $q_0 < {\ln 2}/{8}$, then there exist $0<T<\varepsilon$ such that the ODE system (1.3) has a unique smooth solution (q(t), p(t)) in the interval [0, T) with p(t) increasing, q(t) decreasing, and

$$\begin{aligned} \lim _{t \rightarrow T^{-}} p(t) = \infty \quad \text { and } \quad \lim _{t \rightarrow T^{-}} q(t) = 0. \end{aligned}$$

Using these properties of peakon-antipeakon solutions one can derive the following fundamental estimate on the $H^s$ size of the solution u(t) for $t \in [0, T)$

$$\begin{aligned} \Vert u(t)\Vert _{H^s} \approx \left\{ \begin{array}{ll} p(t)q(t)^{3/2- s},&{}\quad 1/2< s< 3/2, \\ p(t)q(t)\sqrt{\ln (1/q(t))},&{}\quad s = 1/2, \\ p(t)q(t),&{}\quad s < 1/2. \end{array}\right. \end{aligned}$$

Then, using this estimate one proves Proposition 1.1.

Also, we describe the proof of DP ill-posedness for $1/2 \le s < 3/2$, where the continuity of the data-to-solution map fails. Let $u_n(t)$ be the peakon-antipeakon DP solution corresponding to the choice of $\varepsilon =1/n$ and let $u_\infty (t)=0$. Then, by property (2) in norm inflation result we have $\Vert u_n(0)\Vert _{H^s} < 1/n$. So, $u_n(0)$ converges to $u_\infty (0)=0$ in $H^s$. Furthermore, by property (1) the lifespan $T_n$ of each solution $u_n(t)$ satisfies the inequality $T_n<1/n$, whereas the lifespan $T_\infty$ of $u_\infty (t)$ is equal to $\infty$. Since, by property (3) in norm inflation result there is no $T>0$ such that the solutions $u_n(t)$ can be extended to the interval [0, T] for all sufficiently large n we see that continuity condition (iii) of well-posedness fails.

Remark 1.1

Theorem 1.1 about the local well-posedness and nonuniform dependence for CH and DP is also true for the b-family equation (which contains CH when $b=2$ and DP when $b=3$)

$$\begin{aligned} (1-\partial _x^2)u_t = -(b+1)u u_x + b u_x u_{xx} + u u_{xxx} \end{aligned}$$

for $s>3/2$ and all $b\in \mathbb {R}$ (see [21, 27, 30]). For $s <3/2$, in [33] it is shown that the Cauchy problem for the b-family of equations is ill-posed in Sobolev spaces $H^s$ on both the torus and the line when $b > 1$. For $b\le 1$ it remains an open question.

The Euler equations The CH equation apart from being interesting because of its integrability properties and breaking traveling wave solutions, it can serve as a “toy” model for the Euler equations. This is demonstrated by the following result in [46].

Theorem 1.3

The solution map $u_0 \mapsto u$ of the Cauchy problem for the Euler equations

$$\begin{aligned} \partial _{t}u + \nabla _{u}u +\nabla p =0, \quad \textrm{div}\,u = 0, \quad u(0, x)= u_0(x), \end{aligned}$$

is not uniformly continuous from the unit ball in $H^s(\mathbb {T}^n,\mathbb {R}^n)$ into the space $C([0,T]; H^s(\mathbb {T}^n,\mathbb {R}^n))$ for any $s \in \mathbb {R}$, in the periodic case. While, in the non-periodic case it is not uniformly continuous from the unit ball in $H^s(\mathbb {R}^n,\mathbb {R}^n)$ into $C([0,T]; H^s(\mathbb {R}^n,\mathbb {R}^n))$ for any $s>0$.

The proof of Theorem 1.3 is based on well-posedness theory (as it has been developed by Ebin and Marsden [25], Kato and Ponce [54], Majda and Bertozzi [65]) and approximate solutions. Furthermore, it suffices to prove it in two dimensions, since filling the rest of the velocity components with zeros gives the proof in higher dimensions. Here, we shall provide the proof in the periodic case. Motivated by CH, one would think that the following approximate solutions

$$\begin{aligned} u^{\omega ,n}(t,x) = \left( \omega n^{-1} + n^{-s} \cos (nx_2 - \omega t), \, \omega n^{-1} + n^{-s}\cos (n x_1 - \omega t) \right) \end{aligned}$$

is a natural choice. They are divergence free and in one-dimension they collapse to CH approximate solutions. In fact, these are actual solutions! to the Euler equations. The corresponding to $\omega =\pm 1$ sequences $u^{+1,n}(t,x)$ and $u^{-1,n}(t,x)$ satisfy the conditions for nonuniform dependence of the periodic Euler equations in two dimensions. The proof in the non-periodic case is technical since one needs to take care of the errors introduced by the cut-offs that localize the approximate solutions. The localized approximate solutions are not solutions. Bourgain and Li [9] improved this result in the non-periodic case and included the case $s=0$.

The NE and FORQ equations The well-posedness theory developed for the CH equations with quadratic nonlinearities has been extended to the ones with cubic nonlinearities. Next result summarizes this development for NE [35] and FORQ [41].

Theorem 1.4

(Well-posedness for NE and FORQ) If $s>3/2$ in the case of NE and $s>5/2$ in the case of FORQ, then for $u_0 \in H^s$ there exist $T>0$ and a unique solution $u \in C([0,T]; H^s)$ of the initial value problem for NE or FORQ which depends continuously on the initial data $u_0$. Furthermore,

$$\begin{aligned} \Vert u(t)\Vert _{H^s} \le 2 \Vert u_0\Vert _{H^s}\quad \text { for } t \ge \frac{1}{4c_s\Vert u_0\Vert _{H^s}^2} \doteq T, \end{aligned}$$

where $c_s>0$ is a constant depending on s.

(Nonuniform dependence) Also, the data-to-solution map is not uniformly continuous from any bounded subset in $H^s$ into $C([0,T]; H^s)$.

The presence of the cubic nonlinearities in these equations makes the proofs more technical. At the idea level, an important difference from DP and CH is that for NE the approximate solutions used

$$\begin{aligned} u^{\omega , n} = \omega n^{-1/2} + n^{-s} \cos (nx - \omega t)\quad \text { for }\omega = 0,1, \end{aligned}$$

(on the circle) are asymmetric. That is, one of them ($\omega =0$) has no low frequency. Also, the low frequency term in the other is $n^{-1/2}$ instead of $n^{-1}$, which is the case for CH and DP.

Ill-posedness of the Novikov equation For the Novikov equation on both the line and the circle, in [38] the first author, Holliman and Kenig constructed a 2-peakon solution with an asymmetric antipeakon-peakon initial profile whose $H^s$-norm for $s<3/2$ is arbitrarily small. Immediately after the initial time, both the antipeakon and peakon move in the positive direction, and a collision occurs in arbitrarily small time. Moreover, at the collision time the $H^s$-norm of the solution becomes arbitrarily large when $5/4<s<3/2$, thus resulting in norm inflation and ill-posedness. However, when $s<5/4$, the solution at the collision time coincides with a second solitary antipeakon solution. This scenario thus results in nonuniqueness and ill-posedness. Finally, when $s=5/4$ ill-posedness follows either from a failure of convergence or a failure of uniqueness. Considering that the Novikov equation is well-posed for $s>3/2$ [35], these results put together establish 3/2 as the critical index of well-posedness for this equation. This is summarized in the following result.

Theorem 1.5

The Cauchy problem for the Novikov equation on the line and the circle is ill-posed in Sobolev spaces $H^s$ for $s<3/2$. More precisely, if $5/4<s<3/2$ then the data-to-solution map is not continuous while if $s<5/4$ then solution is not unique. When $s=5/4$ then either continuity or uniqueness fails.

Remark 1.2

The case $s=3/2$ has for CH and NE has been studied in [32] where it is shown that the phenomenon of norm inflation occurs which implies ill-posedness when $s=3/2$. For FORQ and $s<3/2$ it is shown in [36] that we have non-uniqueness of solution and thus ill-posedness. The case $3/2\le s\le 5/2$ remains an open question.

Modified Novikov equation Next, we consider the Cauchy problem on the line for the modified by a third order dispersion Novikov equation denoted here by (mNE),

$$\begin{aligned}{} & {} \partial _t u + \kappa \partial ^3_xu + \frac{1}{3} \partial _x(u^3) + (1-\partial _x^2)^{-1}\partial _x \left[ u^3 + \frac{3}{2}u(\partial _x u)^2\right] \nonumber \\{} & {} \qquad \qquad \qquad \qquad \qquad \qquad + (1-\partial _x^2)^{-1}\left[ \frac{1}{2} (\partial _xu)^3 \right] = 0,\end{aligned}$$

(1.4a)

$$\begin{aligned}{} & {} u(x,0)=\varphi (x), \quad t\in {\mathbb {R}}, \quad x\in \mathbb {R}, \end{aligned}$$

(1.4b)

and study its well-posedness in the Sobolev spaces $H^s(\mathbb {R})$. Here, $\kappa >0$ is a parameter. Note that, like in the case of NE, solutions u to this equation conserve their $H^1$ norm. In fact, using the local form of mNE, which results from multiplying mNE by $(1-\partial _x^2)$,

$$\begin{aligned} \partial _{t}u - \partial _{t}\partial _{x}^{2}u + \kappa \partial ^3_xu - \kappa \partial ^5_xu + 4u^2 \partial _x u - 3u\partial _x u \partial _x^2u - u^2 \partial _x^3 u =0, \end{aligned}$$

(1.5)

and doing integration by part we get

$$\begin{aligned} \frac{d}{dt} \int [u^2 + (\partial _x u)^2]dx= & {} 2\int u \cdot \left[ \partial _t u - \partial ^2_x \partial _t u \right] dx\\\overset{(1.5)}{=} & {} 2 \int u \cdot \left[ - \kappa \partial ^3_xu + \kappa \partial ^5_xu - 4u^2\partial _x u + 3u\partial _x u \partial ^2_x u + u^2 \partial ^3_x u\right] dx \\= & {} 2\int \left[ \partial _x\left( u^3 \partial ^2_x u\right) - \partial _x\left( u^4\right) \right] dx = 0. \end{aligned}$$

Next, to state our results for mNE Cauchy problem (1.4) precisely, we recall the definition of Bourgain spaces used in the well-posedness theory of dispersive equations (see, for example, [8, 15, 56]). For any $s,b\in \mathbb {R}$, the Bourgain space $X^{s,b}=X^{s,b}(\mathbb {R}^2)$ is the completion of the Schwartz space $\mathcal {S}(\mathbb {R}^2)$ with respect to the norm

$$\begin{aligned} \Vert u\Vert _{s,b} \doteq \Vert u\Vert _{X^{s,b}} \doteq \left( \int _{\mathbb {R}}\int _{\mathbb {R}}(1+|\xi |)^{2s}(1+|\tau -\xi ^{3}|)^{2b}|\widehat{u}(\xi ,\tau )|^2d\xi d\tau \right) ^{1/2}, \end{aligned}$$

where $\widehat{u}$ denotes the space-time Fourier transform defined by

$$ \widehat{u}(\xi ,\tau )=\int _{\mathbb {R}}\int _{\mathbb {R}}e^{-i(x\xi +t\tau )}u(x,t)dxdt. $$

Also, for $T>0$, $X_T^{s,b}$ denotes the restricted Bourgain space defined by

$$\begin{aligned} X^{s,b}_{T} = \{u : u(x,t) = v(x,t) ~\text { on }~\mathbb {R}\times (-T,T) ~\text { with }~v\in X^{s,b}(\mathbb {R}^2)\}. \end{aligned}$$

(1.6)

Finally, we recall that the norm in $X^{s,b}_{T}$ is defined by

$$\begin{aligned} \Vert u\Vert _{X_T^{s,b}} = \inf _{v\in X^{s,b}}\left\{ \Vert v\Vert _{s,b} : v(x,t) = u(x,t) ~\text { on }~\mathbb {R}\times (-T,T)\right\} . \end{aligned}$$

(1.7)

With the needed definitions in place, next we state our well-posedness result for mNE proved here.

Theorem 1.6

(Local well-posedness) If $s>\frac{2}{3}$, then for any $\varphi \in H^s(\mathbb {R})$ there exist a time $T_{0}=T_0(\Vert \varphi \Vert _{H^s})>0$ and a unique solution u of the Cauchy problem (1.4) on the time interval $[-T_{0},T_{0}]$, such that $u \in X^{s,b}_{T_0} \cap C([-T_{0},T_{0}]; H^s(\mathbb {R}))$, with

$$\begin{aligned} T_{0} = \frac{c_0}{(1+\Vert \varphi \Vert _{H^s}^2)^{\frac{2}{\beta }}}, \quad \beta \doteq \min \left\{ \frac{1}{6}(s-\frac{2}{3}),\frac{1}{8}\right\} , \end{aligned}$$

(1.8)

for some constants $c_{0}=c_{s,b}>0$ depending only on s and b. Moreover, the local solution u satisfies the following size estimate in Bourgain spaces

$$\begin{aligned} \Vert u\Vert _{X_{T_{0}}^{s,b}}\le C(s,b)\Vert \varphi \Vert _{H^s}, \end{aligned}$$

for some $b\in (1/2,1)$. Also, we have the Hadamard space estimate

$$\begin{aligned} \sup _{|t|\le T_0} \Vert u(t)\Vert _{H^s} \le C_0(s, b) \Vert \varphi \Vert _{H^s}. \end{aligned}$$

(1.9)

Finally, the solution depends Lip-continuously on the data $\varphi$.

Considering that mNE solutions conserve their $H^1$ norm, from Theorem 1.6 we get the next result.

Theorem 1.7

(Global well-posedness) The mNE is globally well-posed in the Sobolev space $H^1$.

Theorems 1.6 and 1.7 are motivated by the corresponding results obtained in [13, 42, 43] for the following dispersive modification of the CH equation, denoted by mCH,

$$\begin{aligned} \partial _t u + \kappa \partial ^3_xu + \frac{1}{2} \partial _{x}(u^2) + (1 -\partial _{x}^{2})^{-1} \partial _{x} \left[ u^2 +\frac{1}{2}(\partial _{x}u)^2\right] =0, \end{aligned}$$

which is analogous to the KdV regularization $u_t-6uu_x+\varepsilon ^2u_{xxx}=0$ of the Burgers equation $u_t - 6uu_x=0$, with $\kappa =\varepsilon ^2$, by Lax and Levermore [58,59,60,61]. Although the question of what happens in mCH when the parameter $\kappa$ goes to zero remains open, some numerical results in the periodic case have been obtained in [31]. The above-mentioned work of Lax and Levermore on the small dispersion limit of the Korteweg–de Vries equation provides a good motivation for investigating the corresponding problem for both mCH and mNE. Concerning the proof of well-posedness for mCH, we note that it is based on the KdV bilinear estimate $\Vert \partial _x(f\cdot g)\Vert _{X^{s, b-1}}\le c_s \Vert f\Vert _{X^{s, b}}\Vert g\Vert _{X^{s, b}}$ in [56] that holds for $s>-3/4$ and some $b\in (\frac{1}{2}, 1)$, with f and g replaced with $\partial _xf$ and $\partial _xg$ and therefore is valid for $s>1-\frac{3}{4}=\frac{1}{4}$. This explains why mCH is well-posed for $s>\frac{1}{4}$. Now, considering that the mKdV is well-posed for $s\ge \frac{1}{4}$ [55] and its trilinear estimates hold for $s\ge \frac{1}{4}$ [75], this idea transferred to the mNE situation will give well-posedness for $s\ge 1+\frac{1}{4}=\frac{5}{4}$, which is worse than the well-posedness for $s>\frac{2}{3}$ obtained here. This seems to be an interesting phenomenon for mNE, which is explained by the improved trilinear estimates for the nonlocal nonlinearities stated in the next section. Concerning the mKdV, we mention that its well-posedness in $H^s$ was proved first by Kenig, Ponce and Vega [55] for $s\ge \frac{1}{4}$ with a different method without using its trilinear estimate in Bourgain spaces (also, see [71]). Also, we mention that our proof of the mKdV trilinear estimate provided here for $s>\frac{1}{4}$ is different than the one of Tao in [75]. Concluding, we mention that the sharpness of Theorem 1.6 is an interesting open question. Constructing a counterexample to the nonlocal trilinear estimates (2.18) could provide useful information.

The rest of the paper is organized as follows. In Section 2 we state the three trilinear estimates needed and use them to prove Theorem 1.6. In Section 3 we prove first the trilinear estimate for the local nonlinearity $\partial _x(u^3)$, and then we use it to prove the trilinear estimates for nonlocal nonlinearities $(1-\partial _x^2)^{-1}\partial _x [u(\partial _x u)^2]$ and $(1-\partial _x^2)^{-1}[(\partial _x u)^3]$.

2 Well-posedness in Sobolev spaces—Proof of Theorem 1.6

In what follows we will assume the coefficient $\kappa$ of the dispersion $\partial _x^3u$ is equal to 1, that is $\kappa =1$. When $\kappa >0$ and $\kappa \ne 1$, then the constants appearing in some of the estimates depend on k.

We begin the well-posedness proof by solving the initial value problem (ivp) for the linear mNE

$$\begin{aligned} \partial _t u+\partial _x^{3}u= & {} -w(x, t),\end{aligned}$$

(2.1)

$$\begin{aligned} u(x,0)= & {} \varphi (x), \quad t\in \mathbb {R}, \quad x\in \mathbb {R}, \end{aligned}$$

(2.2)

with forcing w the mNE nonlinearity $w=w_{uuu}$, where

$$\begin{aligned} \begin{aligned} w&= w_{fgh} \doteq \frac{1}{3}\partial _x(f \cdot g \cdot h) + (1-\partial _{x}^{2})^{-1}\partial _{x} \left[ f \cdot g \cdot h +\frac{3}{2}(f \cdot g_x \cdot h_x)\right] \\&\quad + (1-\partial _{x}^{2})^{-1} \left[ \frac{1}{2} (f_x \cdot g_x \cdot h_x)\right] . \end{aligned} \end{aligned}$$

(2.3)

With this definition, we have the following formula to be used later

$$\begin{aligned} \begin{aligned} w_{uuu} -w_{vvv}&= \frac{1}{3} \partial _x[(u-v)(u^2+uv+v^2)] + (1-\partial _x^2)^{-1} \partial _x[(u-v)(u^2+uv+v^2)]\\&\quad + \frac{3}{2} (1-\partial _x^2)^{-1}\partial _x \left[ (u-v)u_x^2 + v(u-v)_x(u+v)_x\right] \\&\quad + \frac{1}{2}(1-\partial _x^2)^{-1} \left[ (u-v)_x(u_x^2 +u_xv_x+v_x^2)\right] . \end{aligned} \end{aligned}$$

(2.4)

We solve initial value problem (ivp) (2.1) by taking Fourier transform with respect to x, which for a test function $\phi (x)$ is defined by the familiar formula

$$\begin{aligned} \widehat{\phi }(\xi )=\int _{\mathbb {R}} e^{-i\xi x}\phi (x) dx, \end{aligned}$$

and solving the resulting DE ivp in the t variable. Thus, we obtain the Duhamel’s formula

$$\begin{aligned} u(x, t) = W(t)\varphi (x) - \int _0^t W(t-t') w(x,t')\, dt', \end{aligned}$$

(2.5)

where $W(t)\varphi (x)$ denotes the solution to the homogeneous Cauchy problem for the linear mNE with initial data $\varphi$, that is

$$\begin{aligned} W(t)\varphi (x) \doteq \frac{1}{2\pi } \int _{-\infty }^\infty e^{i(\xi x+\xi ^{3} t)} \widehat{\varphi }(\xi ) d\xi . \end{aligned}$$

Now, to solve the mNE ivp locally, we introduce the usual time localizer, which is a cut-off function

$$\begin{aligned} \psi \in C_0^\infty (-1, 1), \quad 0\le \psi \le 1, \quad \text { and }\quad \psi (t) = 1 ~\text { for } |t|\le 1/2. \end{aligned}$$

(2.6)

Multiplying the right-hand side (2.5) by $\psi (t)$ we obtain the following global form

$$\begin{aligned} u(x, t) = \psi (t) W(t)\varphi (x) - \psi (t) \int _0^t W(t-t') w (x, t')\, dt' \doteq {\varPhi } u(x, t). \end{aligned}$$

(2.7)

Note that for $|t|\le 1/2$ global formulation (2.7) coincides with local formulation (2.5). Thus, our strategy for proving existence of a local solution of our ivp is to show that the map ${\varPhi }$ defined in (2.7) has a fixed point in appropriate solution space, via a fixed point argument. For this, we assume that the forcing w is globally defined and using the relation $\widehat{w}^x(\xi , t') = \frac{1}{2\pi } \int _{-\infty }^\infty e^{i\tau t'} \widehat{w} (\xi , \tau )d\tau$ we obtain the following decomposition of the map ${\varPhi }$, like in [8],

$$\begin{aligned} \!\!\!\!\!\!\!\!\!\!\!\!{\varPhi } u(x, t)= & {} \frac{1}{2\pi } \psi (t) \int _{-\infty }^\infty e^{i(\xi x+\xi ^{3} t)} \widehat{\varphi }(\xi ) d\xi \end{aligned}$$

(2.8)

$$\begin{aligned}{} & {} +\frac{i}{4\pi ^2} \psi (t) \int _{\mathbb {R}}\int _{\mathbb {R}} e^{i(\xi x+\tau t)} \frac{1-\psi (\tau -\xi ^{3})}{\tau -\xi ^{3}} \widehat{w}(\xi ,\tau ) d\tau d\xi \end{aligned}$$

(2.9)

$$\begin{aligned}{} & {} - \frac{i}{4\pi ^2} \psi (t) \int _{\mathbb {R}}\int _{\mathbb {R}} e^{i(\xi x+\xi ^{3}t)} \frac{1-\psi (\tau -\xi ^{3})}{\tau -\xi ^{3}} \widehat{w}(\xi ,\tau ) d\tau d\xi \end{aligned}$$

(2.10)

$$\begin{aligned}{} & {} + \frac{i}{4\pi ^2} \psi (t) \int _{\mathbb {R}}\int _{\mathbb {R}} e^{i(\xi x+\xi ^{3} t)} \frac{\psi (\tau -\xi ^{3})[e^{i(\tau - \xi ^{3})t} - 1]}{\tau - \xi ^{3}} \widehat{w}(\xi ,\tau ) d\tau d\xi . \end{aligned}$$

(2.11)

Next, using this convenient for estimation in Bourgain spaces form of the iteration map ${\varPhi }$, we show that it is a contraction in an appropriate ball of the Bourgain $X^{s, b}$ for $s>\frac{2}{3}$ and some $b\in (\frac{1}{2}, 1)$. For proving that the map ${\varPhi }$ is onto we begin with the inequality

$$\begin{aligned} \Vert {\varPhi } u\Vert _{X^{s,b}} \le \Vert \psi (t) W(t)\varphi (x)\Vert _{s,b} + \left\| \psi (t) \int _0^t W(t-t') w (x, t')\, dt')\right\| _{s,b}, \quad w=w_{uuu}, \end{aligned}$$

(2.12)

where $w_{uuu}$ is given by formula (2.3), while to show that ${\varPhi }$ is a contraction we begin with inequality

$$\begin{aligned} \Vert {\varPhi } u -{\varPhi } v\Vert _{X^{s,b}} \le \left\| \psi (t) \int _0^t W(t-t') w(x, t')\,dt')\right\| _{s,b}, \quad w=w_{uuu}-w_{vvv}, \end{aligned}$$

(2.13)

where $w_{uuu}-w_{vvv}$ is given by formula (2.4). From inequalities (2.12) and (2.13) we see that we need the following linear estimates, whose proof follows from using the convenient writing (2.8)–(2.11) of ${\varPhi } u$, and the definition of $X^{s,b}$-norm.

Lemma 2.1

(Linear estimates) For any $s\in \mathbb {R}$ and $b>0$ there is $c_1=c_1(\psi ,s,b)$ such that

$$\begin{aligned} \Vert \psi (t) W(t)\varphi (x)\Vert _{s,b} \le c_1 \Vert \varphi \Vert _{H^s}. \end{aligned}$$

(2.14)

Also, for $s\in \mathbb {R}$ and $1/2<b<1$ we have

$$\begin{aligned} \left\| \psi (t) \int _0^t W(t-t') w (x,t')\,dt'\right\| _{s,b} \le c_1 \Vert w\Vert _{s,b-1}. \end{aligned}$$

(2.15)

Now, taking into consideration the form of mNE nonlinearities (2.3) we see that the following three trilinear estimates are the key ingredients for showing that our iteration map is a contraction, thus proving Theorem 1.6.

Proposition 2.1

(mKdV trilinear estimate) If $s> \frac{1}{4}$, then there exist b and $b'$ with $1/2<b' \le b<1$ such that the following trilinear estimate holds:

$$\begin{aligned} \Vert \partial _x(fgh) \Vert _{X^{s,b-1}} \le c_2\Vert f\Vert _{X^{s,b'}} \Vert g\Vert _{X^{s,b'}}\Vert h\Vert _{X^{s,b'}}, \end{aligned}$$

(2.16)

where $c_2=c_2(s,b)$. In fact, b and $b'$ depend on s and can be chosen as follows:

$$\begin{aligned} \frac{1}{2}<b' \le b \le \frac{1}{2} + \beta _1, \quad \text {where }\beta _1 = \beta _1(s) \doteq \min \left\{ \frac{1}{6}\left( s-\frac{1}{4}\right) , \frac{1}{8}\right\} . \end{aligned}$$

Proposition 2.2

(First nonlocal trilinear estimate) If $s> \frac{1}{2}$, then there exist b and $b'$ with $1/2<b' \le b<1$ such that the following trilinear estimate holds:

$$\begin{aligned} \Vert (1+\partial _x^2)^{-1}\partial _x[f(\partial _xg)(\partial _xh)]\Vert _{X^{s,b-1}} \le c_2\Vert f\Vert _{X^{s,b'}}\Vert g\Vert _{X^{s,b'}}\Vert h\Vert _{X^{s,b'}}, \end{aligned}$$

(2.17)

where $c_2=c_2(s,b)$. Also, b and $b'$ depend on s and can be chosen as follows:

$$\begin{aligned} \frac{1}{2} < b' \le b \le \frac{1}{2} + \beta _2, \quad \text { where }\beta _2 = \beta _2(s) \doteq \min \left\{ \frac{1}{6}\left( s-\frac{1}{2}\right) ,\frac{1}{8}\right\} . \end{aligned}$$

Proposition 2.3

(Second nonlocal trilinear estimate) If $s> \frac{2}{3}$, then there exist b and $b'$ with $1/2<b' \le b<1$ such that the following trilinear estimates holds:

$$\begin{aligned} \Vert (1+\partial _x^2)^{-1}[(\partial _xf)(\partial _xg)(\partial _xh)]\Vert _{X^{s,b-1}} \le c_2\Vert f\Vert _{X^{s,b'}}\Vert g\Vert _{X^{s,b'}}\Vert h\Vert _{X^{s,b'}}, \end{aligned}$$

(2.18)

where $c_2=c_2(s,b)$. Finally, b and $b'$ depend on s and can be chosen as follows:

$$\begin{aligned} \frac{1}{2} < b' \le b \le \frac{1}{2} + \beta _3,\quad \text { where } \beta _3 = \beta _3(s) \doteq \min \left\{ \frac{1}{6}\left( s-\frac{2}{3}\right) , \frac{1}{8}\right\} . \end{aligned}$$

Remark 2.1

Note that $\beta _3$ is the smallest of all parameters $\beta _j$ used in the three trilinear estimates above. Therefore, all these estimates hold for

$$\begin{aligned} \frac{1}{2} < b' \le b \le \frac{1}{2} + \beta , \quad \text { where } \beta = \beta (s) \doteq \min \left\{ \frac{1}{6}\left( s-\frac{2}{3}\right) , \frac{1}{8}\right\} , \end{aligned}$$

and this is what we will use in the proof of our well-posedness Theorem 1.6.

As we have mentioned earlier, our proof of the mKdV trilinear estimate (2.16) is different than the one given in [75]. Also, a good part of the proof of each one of the nonlocal trilinear estimates (2.17) and (2.18) is reduced to the proof of estimate (2.16).

We shall prove the trilinear estimates in the next section. Here, we shall use them to complete the proof of Theorem 1.6. For any size initial data $\varphi \in H^s$ and lifespan T such that

$$\begin{aligned} 0<T<1/2, \end{aligned}$$

and to be determined later, we further localize integral equation (2.7) as follows

$$\begin{aligned} \begin{aligned} u(x, t)&= \psi (t) W(t)\varphi - \psi (t) \int _0^t W(t-t') \psi _{2T}(t') w_{uuu} (x, t')\, dt'\\&\doteq {\varPhi }_T (u) (x, t) = {\varPhi }_{T,\varphi }(u)(x, t), \end{aligned} \end{aligned}$$

(2.19)

where $\psi _T(t)=\psi (t/T)$, with $\psi$ being the standard localization function defined in (2.6). If $|t| <T$, then we see that ${\varPhi }_T(u)={\varPhi }(u)$ and the fixed point of the iteration map (2.19) is the solution to the modified Novikov equation ivp (1.4). Next, we show that the iteration map (2.19) is contraction on the ball

$$\begin{aligned} B(r) = \{u\in X^{s,b}:\Vert u\Vert _{X^{s,b}} \le r\}, \end{aligned}$$

if we choose the radius r and the lifespan T appropriately. For this we will need following multiplier estimate in [74] (see Lemma 2.11).

Lemma 2.2

Let $\eta (t)$ be a function in the Schwartz space $\mathcal {S}(\mathbb {R})$. If $-\frac{1}{2}< \gamma ' \le \gamma < \frac{1}{2}$, then for any $0<T\le 1$ we have

$$\begin{aligned} \Vert \eta (t/T)u\Vert _{X^{s,\gamma '}} \le c_3(\eta ,\gamma ,\gamma ') T^{\gamma -\gamma '} \Vert u\Vert _{X^{s,\gamma }}. \end{aligned}$$

(2.20)

Also, we will use the trilinear estimates (2.16), (2.17), and (2.18), with the following choice of $b'$ and b

$$\begin{aligned} b = \frac{1}{2} + \frac{1}{2}\beta ~\text { (in place of b')} \quad \text { and } \quad b_1 = \frac{1}{2}+\beta ~\text { (in place of { b})},~~\text { with }\beta =\beta _3, \end{aligned}$$

in which they read as follows

$$\begin{aligned} \Vert \partial _x(fgh)\Vert _{s,b_1-1}\le & {} c_2 \Vert f\Vert _{s,b} \Vert g\Vert _{s,b}\Vert h\Vert _{s,b},\end{aligned}$$

(2.21a)

$$\begin{aligned} \Vert (1+\partial _x^2)^{-1}\partial _x[f(\partial _xg)(\partial _xh)]\Vert _{s,b_1-1}\le & {} c_2\Vert f \Vert _{s,b} \Vert g \Vert _{s,b}\Vert h \Vert _{s,b},\end{aligned}$$

(2.21b)

$$\begin{aligned} \Vert (1+\partial _x^2)^{-1}[(\partial _xf)(\partial _xg)(\partial _xh)]\Vert _{s,b_1-1}\le & {} c_2\Vert f \Vert _{s,b} \Vert g \Vert _{s,b}\Vert h \Vert _{s,b}. \end{aligned}$$

(2.21c)

${\varPhi }_ T$ is onto: For $u\in B(r)$, applying the linear estimates (2.14), (2.15) and the multiplier estimate (2.20) with $\gamma =b_1-1$, $\gamma '=b-1$, we get

$$\begin{aligned} \begin{aligned} \Vert {\varPhi }_ {T}(u)\Vert _{s,b}&\le c_1\Vert \varphi \Vert _{H^s} + c_1\left\| \psi _{2T}(t)w_{uuu}\right\| _{s,b-1}\\&\le c_1\Vert \varphi \Vert _{H^s} + c_1 c_3T^{\frac{1}{2}\beta }\left\| w_{uuu}\right\| _{s,b_1-1}, \end{aligned} \end{aligned}$$

(2.22)

since $b_1-b=\frac{\beta }{2}$. Then, applying trilinear estimates (2.21) to the nonlinearity $w_{uuu}$ in (2.3), we get

$$\begin{aligned} \Vert w_{uuu}\Vert _{s,b_1-1}= & {} \left\| \frac{1}{3} \partial _x(u^3) + (1 - \partial _x^2)^{-1} \partial _x\left[ u^3 + \frac{3}{2}u(\partial _x u)^2\right] + (1 - \partial _x^2)^{-1}\left[ \frac{1}{2} (\partial _xu)^3\right] \right\| _{s,b-1}\nonumber \\\le & {} 4c_2 \Vert u\Vert ^3_{s,b}. \end{aligned}$$

(2.23)

Thus, combining (2.22) and (2.23) gives the onto estimate

$$\begin{aligned} \Vert {\varPhi }_ {T}(u)\Vert _{s,b} \le c_1\Vert \varphi \Vert _{H^s} + 4c_1c_2c_3 T^{\frac{1}{2}\beta }\Vert u\Vert ^3_{s,b}. \end{aligned}$$

(2.24)

From (2.24) we see that for the map ${\varPhi }_T$ to be onto, it suffices to have $c_1\Vert \varphi \Vert _{H^s} + 4c_1c_2c_3T^{\frac{1}{2}\beta }\Vert u\Vert ^3_{s,b} \le r$. And, since $u\in B(r)$ it suffices for the lifespan T and the radius r to satisfy the condition

$$\begin{aligned} c_1\Vert \varphi \Vert _{H^s} + 4c_1c_2c_3T^{\frac{1}{2}\beta } r^3 \le r. \end{aligned}$$

(2.25)

${\varPhi }_ T$ is contraction For $u,v\in B(r)$, applying the linear estimate (2.15) and the multiplier estimate (2.20) with $\gamma =b_1-1$, $\gamma '=b-1$, we have

$$\begin{aligned} \begin{aligned} \Vert {\varPhi }_{T}(u)-{\varPhi }_{T}(v)\Vert _{s,b}&\le c_1\Vert \psi _{2T}(t)(w_{uuu}-w_{vvv})\Vert _{s,b-1}\\&\le c_1 c_3T^{\frac{1}{2}\beta }\Vert w_{uuu}-w_{vvv}\Vert _{s,b_1-1}. \end{aligned} \end{aligned}$$

(2.26)

Next, applying trilinear estimates (2.21) to the nonlinearities $w_{uuu}-w_{vvv}$ defined by (2.4), we get

$$\begin{aligned} \Vert w_{uuu}-w_{vvv}\Vert _{s,b_1-1}\le & {} \frac{4}{3} c_2\Vert u-v\Vert _{s,b} \left( \Vert u\Vert _{s,b}^2 +\Vert u\Vert _{s,b}\Vert v\Vert _{s,b} +\Vert v\Vert _{s,b}^2\right) \\{} & {} + \frac{3}{2} c_2 \Vert u-v\Vert _{s,b} \left( \Vert u\Vert _{s,b}^2 +\Vert v\Vert _{s,b} \Vert u+v\Vert _{s,b}\right) \\{} & {} + \frac{1}{2} c_2 \Vert u-v\Vert _{s,b} \left( \Vert u\Vert _{s,b}^2 +\Vert u\Vert _{s,b}\Vert v\Vert _{s,b} +\Vert v\Vert _{s,b}^2\right) \\\le & {} 4c_2 \left( \Vert u\Vert _{s,b}^2 +\Vert u\Vert _{s,b}\Vert v\Vert _{s,b} +\Vert v\Vert _{s,b}^2\right) \Vert u-v\Vert _{s,b}, \end{aligned}$$

which combined with (2.26) gives the contraction estimate

$$\begin{aligned} \Vert {\varPhi }_{T}(u)-{\varPhi }_{T}(v)\Vert _{s,b} \le 12c_1c_2c_3r^2 T^{\frac{1}{2}\beta } \Vert u-v\Vert _{s,b}. \end{aligned}$$

Thus, in order to make the iteration map ${\varPhi }_ T$ a contraction map, it suffices to have

$$\begin{aligned} 12c_1c_2c_3r^2T^{\frac{1}{2}\beta } \le \frac{1}{2}. \end{aligned}$$

(2.27)

Combining conditions (2.25) with (2.27), we see that it suffices to have $c_1\Vert \varphi \Vert _{H^s} + \frac{1}{6}r \le r$ or $r\ge \frac{6}{5} c_1\Vert \varphi \Vert _{H^s}$. So, we choose the radius to be

$$\begin{aligned} r \doteq 2c_1\Vert \varphi \Vert _{H^s}. \end{aligned}$$

(2.28)

Then, from (2.27), it suffices to have $T^{\frac{1}{2}\beta } \le (24 c_1c_2c_3r^2)^{-1}$ or $T \le (24 c_1c_2c_3r^2)^{-2/\beta }$, which follows from choosing $T = \frac{1}{2} \frac{1}{(1+24c_1c_2c_3r^2)^{\frac{2}{\beta }}}<\frac{1}{2}$. Combining this choice of T together with choice (2.28) for r gives

$$\begin{aligned} T = \frac{1}{2} \frac{1}{(1+96c_1^3c_2c_3\Vert \varphi \Vert _{H^s}^2)^{\frac{2}{\beta }}} \ge \frac{c_0}{(1+\Vert \varphi \Vert _{H^s}^2)^{\frac{2}{\beta }}} \doteq T_0, \end{aligned}$$

(2.29)

for some $c_0$ depending on $c_1, c_2, c_3$ and $\beta$, that is $c_0=c_0(s,b)$. This is the estimate for the lifespan (1.8) stated in Theorem 1.6.

Now, observe that the fixed-point $u\in X^{s,b}(\mathbb {R}^2)$ of the iteration map (2.19) restricted to $|t|\le T_0$ satisfies the non-$\psi$ integral equation (2.5), that is

$$\begin{aligned} u(x,t) = W(t)\varphi (x) - \int _0^t W(t-t') w (x, t')\, dt', \quad |t|\le T_0. \end{aligned}$$

Therefore, we have a solution u for the modified Novikov equation ivp (1.4a) in the space $X^{s,b}_{T_0}$ defined by (1.6), and having norm $\Vert u\Vert _{X_T^{s,b}}$ defined by (1.7).

Solution Bound Since the solution $u\in X^{s,b}_{T_0}$ is the restriction of the fixed-point $v(x,t)={\varPhi }_{T_0}(v)$ (which in our discussion above we were calling u), we have

$$\begin{aligned} \Vert u\Vert _{X_{T_0}^{s,b}} \le \Vert v\Vert _{s,b} = \Vert {\varPhi }_{T_0}(v)\Vert _{s,b} \le 2c_1\Vert \varphi \Vert _{H^s}, \end{aligned}$$

where the last step follows from fact that v is a fixed-point in the ball $B(2c_1\Vert \varphi \Vert _{H^s})$. Finally, in order to prove estimate (1.9) we need the following basic result, which follows the definition of $\Vert \cdot \Vert _{H^s}$ and the Cauchy–Schwarz inequality.

Lemma 2.3

Let $b>\frac{1}{2}$. The inclusion $X^{s,b}(\mathbb {R}^2)\hookrightarrow C(\mathbb {R},H^{s}(\mathbb {R}))$ is continuous, that is

$$\begin{aligned} \sup _{t\in \mathbb {R}} \Vert u(t)\Vert _{H^{s}} \le C(b) \Vert u\Vert _{X^{s,b}}. \end{aligned}$$

Now, the desired solution bound follows from Lemma 2.3 since

$$ \sup _{|t|\le T_0} \Vert u(t)\Vert _{H^s} = \sup _{|t|\le T_0} \Vert v(t)\Vert _{H^s} \le \sup _{t\in \mathbb {R}}\Vert v(t)\Vert _{H^s} \le C\Vert v\Vert _{X^{s,b}} \le 2c_1C\Vert \varphi \Vert _{H^s}. $$

Lip-continuous dependence on initial data For any $\varphi _0\in H^s$ and $R>0$ let us consider the neighborhood

$$\begin{aligned} U = B(\varphi _0, R) \doteq \{\varphi \in H^s : \Vert \varphi -\varphi _0\Vert _{H^s} \le R\}. \end{aligned}$$

Then, by the triangle inequality we have $\Vert \varphi \Vert _{H^s} \le \Vert \varphi -\varphi _0\Vert _{H^s} + \Vert \varphi _0\Vert _{H^s} = \Vert \varphi _0\Vert _{H^s}+R$. Using this and lifespan estimate (2.29) we see that

$$\begin{aligned} T^*\doteq \frac{1}{2} \frac{1}{[1+96c_1^3c_2c_3(\Vert \varphi _0\Vert _{H^s}+R)^2]^{\frac{2}{\beta }}} \end{aligned}$$

(2.30)

is common lifespan of solutions to modified Novikov equation ivp (1.4) with initial data in U. Now, for $\varphi _k\in U$, $k=1,2$, we denote the solution of the modified Novikov equation ivp (1.4a) with the initial data $\varphi _k$ by $u_k$, that is $u_k$ satisfies

$$\begin{aligned} u_k(x,t) \doteq \psi (t) W(t)\varphi _k - \psi (t) \int _0^t W(t-t') \psi _{2T^*}(t') w_{u_ku_ku_k} dt', \quad |t| < T^*. \end{aligned}$$

Moreover, we denote the fixed point of iteration map ${\varPhi }_{T^*,\varphi _k}(v_k)$ by $v_k$, that is

$$\begin{aligned} v_k(x,t) \doteq \psi (t) W(t)\varphi _k - \psi (t)\int _0^t W(t-t') \psi _{2T^*}(t') w_{v_kv_kv_k} dt', \quad t\in \mathbb {R}. \end{aligned}$$

Since $v_k$ is extensions of $u_k$ from $\mathbb {R}\times [-T^*, T^*]$ to $\mathbb {R}\times \mathbb {R}$, we have $\Vert u_1-u_2\Vert _{X_{T^*}^{s,b}} \le \Vert v_1-v_2\Vert _{s,b}$. Thus, to prove the Lip-continuous dependence on initial data, i.e. $\Vert u_1-u_2\Vert _{X_{T^*}^{s,b}} \le C\Vert \varphi _1- \varphi _2\Vert _{H^s}$, where C is a constant depending on $\varphi _k$, it suffices to show that

$$\begin{aligned} \Vert v_1-v_2\Vert _{s,b} \le C \Vert \varphi _1-\varphi _2\Vert _{H^s}. \end{aligned}$$

(2.31)

Now, as before, we choose the balls

$$\begin{aligned} B_k = B_k(2c_1\Vert \varphi _k\Vert _{H^s}) \doteq \left\{ v\in X^{s,b}: \Vert v\Vert _{s,b} \le 2c_1\Vert \varphi _k\Vert _{H^s}\right\} ,\quad k=1,2. \end{aligned}$$

(2.32)

From the proof of existence, we know that $v_k\in B_k$. So, applying the linear estimates (2.14), (2.15) and the multiplier estimate (2.20) with $\gamma =b_1-1$, $\gamma '=b-1$, we get

$$\begin{aligned} \Vert v_1-v_2\Vert _{s,b}\le & {} c_1\Vert \varphi _1-\varphi _2\Vert _{H^s} + c_1 \Vert \psi _{2T^*}(t)(w_{v_1v_1v_1}-w_{v_2v_2v_2})\Vert _{s,b-1}\\\le & {} c_1\Vert \varphi _1-\varphi _2\Vert _{H^s} + c_1c_3T^{*\frac{1}{2}\beta }\Vert w_{v_1v_1v_1}-w_{v_2v_2v_2}\Vert _{s,b_1-1}. \end{aligned}$$

Then, applying trilinear estimates (2.21a2.21b2.21c) to the nonlinearities $w_{v_1v_1v_1}-w_{v_2v_2v_2}$ defined by (2.4), we get

$$\begin{aligned} \begin{aligned} \Vert w_{v_1v_1v_1}-w_{v_2v_2v_2}\Vert _{s,b_1-1}&\le \frac{4}{3} c_2 \Vert v_1-v_2\Vert _{s,b} \left( \Vert v_1\Vert _{s,b}^2 +\Vert v_1\Vert _{s,b}\Vert v_2\Vert _{s,b} +\Vert v_2\Vert _{s,b}^2\right) \\&\quad + \frac{3}{2} c_2 \Vert v_1-v_2\Vert _{s,b} \left( \Vert v_1\Vert _{s,b}^2 +\Vert v_1\Vert _{s,b} \Vert v_1+v_2\Vert _{s,b}\right) \\&\quad + \frac{1}{2} c_2 \Vert v_1-v_2\Vert _{s,b} \left( \Vert v_1\Vert _{s,b}^2 +\Vert v_1\Vert _{s,b}\Vert v_2\Vert _{s,b} +\Vert v_2\Vert _{s,b}^2\right) \\&\le 4c_2 \left( \Vert v_1\Vert _{s,b}^2 +\Vert v_1\Vert _{s,b}\Vert v_2\Vert _{s,b} +\Vert v_2\Vert _{s,b}^2\right) \Vert v_1-v_2\Vert _{s,b},\qquad \end{aligned} \end{aligned}$$

(2.33)

which combined with definition (2.32) and estimate (2.33) implies that

$$\begin{aligned} \Vert v_1-v_2\Vert _{s,b} \le c_1\Vert \varphi _1-\varphi _2\Vert _{H^s} + 48c_1^3 c_2c_3 T^{*\frac{1}{2}\beta } \cdot (\Vert \varphi _0\Vert _{H^s}+R)^2 \Vert v_1-v_2\Vert _{s,b}. \end{aligned}$$

Combining this with the expression (2.30) for the common lifespan $T^*$, we obtain

$$\begin{aligned} \Vert v_1-v_2\Vert _{s,b}\le & {} c\Vert \varphi _1-\varphi _2\Vert _{H^s} + \frac{48c_1^3c_2c_3 (\Vert \varphi _0\Vert _{H^s}+R)^2}{1+96c_1^3c_2c_3(\Vert \varphi _0\Vert _{H^s}+R)^2}\Vert v_1-v_2\Vert _{s,b}\\\le & {} c\Vert \varphi _1-\varphi _2\Vert _{H^s} + \frac{1}{2}\Vert v_1-v_2\Vert _{s,b}, \end{aligned}$$

which is the desired estimate (2.31). Now, using Lemma 2.3, we can also get the estimate in the space $C(\mathbb {R};H^s)$, i.e.

$$ \sup _{t\in \mathbb {R}} \Vert v_1(t)-v_2(t)\Vert _{H^s} \le \Vert v_1-v_2\Vert _{s,b} \le 2c_1C\Vert \varphi _1-\varphi _2\Vert _{H^s}. $$

Finally, concerning the proof of uniqueness of solution in the space $X_{T_0}^{s,b}$, it is similar to one presented in [6]. This completes the proof of Theorem 1.6. $\square$

3 Proof of Trilinear Estimates in Bourgain Spaces

In this section, we prove the trilinear estimates (2.16), (2.17) and (2.18). For this, we need the following calculus estimates [51, 56].

Lemma 3.1

If $\ell >1/2$ and $\ell '>\frac{1}{2}$ then

$$\begin{aligned} \int _{\mathbb {R}}\frac{dx}{(1+|x-a|)^{2\ell }(1+|x-c|)^{2\ell }}\lesssim & {} \frac{1}{(1+|a-c|)^{2\ell }},\end{aligned}$$

(3.1)

$$\begin{aligned} \int _{\mathbb {R}}\frac{dx}{(1+|x-a|)^{2(1-\ell )}(1+|x-c|)^{2\ell '}}\lesssim & {} \frac{1}{(1+|a-c|)^{2(1-\ell )}}. \end{aligned}$$

(3.2)

In addition, if $\frac{1}{4}< \ell ,\ell '<\frac{1}{2}$, then

$$\begin{aligned} \int _{\mathbb {R}}\frac{dx}{(1+|x-a|)^{2\ell }(1+|x-c|)^{2\ell '}}\lesssim \frac{1}{(1+|a-c|)^{2\ell +2\ell '-1}}. \end{aligned}$$

(3.3)

3.1 Proof of mKdV Trilinear Estimate (2.16)

We start with expressing trilinear estimate (2.16), i.e. $\Vert \partial _x(fgh)\Vert _{X^{s,b-1}}\! \le \! c_{s,b}\Vert f\Vert _{X^{s,b'}}\Vert g\Vert _{X^{s,b'}}$ $\Vert h\Vert _{X^{s,b'}}$, in its $L^2$ form. Using the following notation

$$\begin{aligned} c_u \doteq |\widehat{u}(\xi ,\tau )| (1+|\xi |)^{s} (1+|\tau -\xi ^3|)^{b'}, \end{aligned}$$

(3.4)

we get $\Vert u\Vert _{X^{s,b'}} = \Vert c_u\Vert _{L^2}$. Next, we form the $X^{s,b-1}$-norm, that is

$$\begin{aligned}{} & {} \Vert \partial _x(fgh)\Vert _{X^{s,b-1}}^2\\= & {} \int _{\mathbb {R}^2} \frac{|\xi |^2(1+|\xi |)^{2s}}{(1+|\tau -\xi ^3|)^{2(1-b)}}\\{} & {} \times \left[ \int _{\mathbb {R}^4} \widehat{f} (\xi _1,\tau _1) \widehat{g} (\xi _2,\tau _2) \widehat{h} (\xi -\xi _1-\xi _2,\tau -\tau _1-\tau _2) d\xi _2d\tau _2 d\xi _1d\tau _1\right] ^2d\xi d\tau \\\le & {} \int _{\mathbb {R}^2} \frac{|\xi |^2(1+|\xi |)^{2s}}{(1+|\tau -\xi ^3|)^{2(1-b)}}\\{} & {} \times \left( \int _{\mathbb {R}^4} \left| \widehat{f}(\xi _1,\tau _1)\widehat{g}(\xi _2,\tau _2) \widehat{h} (\xi -\xi _1-\xi _2,\tau -\tau _1-\tau _2)\right| d\xi _2d\tau _2 d\xi _1d\tau _1\right) ^2d\xi d\tau \\= & {} \int _{\mathbb {R}^2} \left( \int _{\mathbb {R}^2}\int _{\mathbb {R}^2} Q \cdot c_f(\xi _1,\tau _1) c_g(\xi _2,\tau _2) c_h(\xi -\xi _1-\xi _2,\tau -\tau _1-\tau _2) d\xi _2d\tau _2 d\xi _1d\tau _1\right) ^2 d\xi d\tau , \end{aligned}$$

where Q is defined in (3.6) below. Thus, to prove trilinear estimate (2.16), it suffices to show the $L^2$ inequality

$$\begin{aligned} \begin{aligned}&\left\| \int _{\mathbb {R}^4} Q \cdot c_f(\xi _1,\tau _1) c_g(\xi _2,\tau _2) c_h(\xi -\xi _1-\xi _2,\tau -\tau _1-\tau _2) d\xi _2d\tau _2 d\xi _1d\tau _1\right\| _{L^2_{\xi ,\tau }}^2\\&\lesssim \Vert c_f\Vert _{L^2_{\xi ,\tau }}^2\Vert c_g\Vert _{L^2_{\xi ,\tau }}^2\Vert c_h\Vert _{L^2_{\xi ,\tau }}^2, \end{aligned} \end{aligned}$$

(3.5)

where the multiplier $Q=Q(\xi ,\tau ,\xi _1,\tau _1,\xi _2,\tau _2)$ is defined by

$$\begin{aligned} Q \doteq \frac{|\xi |(1+|\xi |)^{s}(1+|\xi _1|)^{-s}(1+|\xi _2|)^{-s}(1+|\xi -\xi _1-\xi _2|)^{-s}}{(1+|\tau -\xi ^3|)^{1-b}(1+|\tau _1-\xi _1^3|)^{b'}(1+|\tau _2-\xi _2^3|)^{b'}(1+|\tau -\tau _1-\tau _2-(\xi -\xi _1-\xi _2)^3|)^{b'}}. \end{aligned}$$

(3.6)

In the multiplier Q we recognize the familiar Bourgain quantity

$$\begin{aligned} \begin{aligned} d_3(\xi ,\xi _1,\xi _2)&\doteq (\tau -\xi ^3)-(\tau _1-\xi _1^3)-(\tau _2-\xi _2^3)-[\tau -\tau _1-\tau _2-(\xi -\xi _1-\xi _2)^3]\\&= -\xi ^3+\xi _1^3+\xi _2^3+(\xi -\xi _1-\xi _2)^3. \end{aligned} \end{aligned}$$

(3.7)

Below, we list two useful and elementary properties for this quantity.

Lemma 3.2

The Bourgain quantity $d_3 (\xi ,\xi _1,\xi _2)$ satisfies the following properties:

$$\begin{aligned} d_3 (\xi ,\xi _1,\xi _2)= & {} -3(\xi -\xi _1)(\xi -\xi _2)(\xi _1+\xi _2),\end{aligned}$$

(3.8)

$$\begin{aligned} \frac{\partial d_3}{\partial \xi _1}= & {} 3\xi _1^2-3(\xi -\xi _1-\xi _2)^2 = 3(\xi -\xi _2)(2\xi _1-\xi +\xi _2)\nonumber \\= & {} 6(\xi -\xi _2)\left[ \xi _1-\frac{1}{2}(\xi -\xi _2)\right] . \end{aligned}$$

(3.9)

Now, we estimate the left-hand side of estimate (3.5) in a way similar to the case of trilinear estimates for the cubic nonlinear Schrödinger equation [26]. First applying the Cauchy–Schwarz inequality in $\xi _1$, $\xi _2$, $\tau _1$, $\tau _2$, and then using Hölder’s inequality we get

$$\begin{aligned}{} & {} \left\| \int _{\mathbb {R}^4} Q \cdot c_f(\xi _1,\tau _1) c_g(\xi _2,\tau _2) c_h(\xi -\xi _1-\xi _2,\tau -\tau _1-\tau _2)d\xi _2d\tau _2d\xi _1d\tau _1\right\| _{L^2_{\xi ,\tau }}^2\nonumber \\\le & {} \int _{\mathbb {R}^2} \left( \int _{\mathbb {R}^4} Q^2 d\xi _2d\tau _2 d\xi _1d\tau _1\right) \nonumber \\{} & {} \times \left( \int _{\mathbb {R}^4} c_f^2(\xi _1,\tau _1) c_g^2(\xi _2,\tau _2) c_h^2(\xi -\xi _1-\xi _2,\tau -\tau _1-\tau _2) d\xi _2d\tau _2 d\xi _1d\tau _1\right) d\xi d\tau \nonumber \\= & {} \int _{\mathbb {R}^2} \left( \int _{\mathbb {R}^4} Q^2 d\xi _2d\tau _2 d\xi _1d\tau _1\right) \left( c_f^2 *c_g^2 *c_h^2\right) (\xi ,\tau ) d\xi d\tau \nonumber \\\le & {} \left\| \int _{\mathbb {R}^4} Q^2 d\xi _2d\tau _2 d\xi _1d\tau _1 \right\| _{L_{\xi ,\tau }^\infty }\left\| c_f^2 *c_g^2 *c_h^2\right\| _{L_{\xi ,\tau }^1}. \end{aligned}$$

(3.10)

Furthermore, applying Young’s convolution inequality $\Vert f*g\Vert _{L^1(\mathbb {R})}\le \Vert f\Vert _{L^1(\mathbb {R})}\Vert g\Vert _{L^1(\mathbb {R})}$ twice, we get

$$\begin{aligned} \begin{array}{lll} \Vert c_f^2 *c_g^2 *c_h^2\Vert _{L_{\xi ,\tau }^1} &{}\le &{} \Vert c_f^2 *c_g^2\Vert _{L_{\xi ,\tau }^1} \cdot \Vert c_h^2\Vert _{L_{\xi ,\tau }^1}\\ &{}\le &{} \Vert c_f^2\Vert _{L_{\xi ,\tau }^1} \Vert c_g^2\Vert _{L_{\xi ,\tau }^1}\Vert c_h^2\Vert _{L_{\xi ,\tau }^1}= \Vert c_f\Vert _{L^2_{\xi ,\tau }}^2 \Vert c_g\Vert _{L^2_{\xi ,\tau }}^2\Vert c_h\Vert _{L^2_{\xi ,\tau }}^2. \end{array} \end{aligned}$$

(3.11)

Combining estimate (3.11) with (3.10), we get

$$\begin{aligned}{} & {} \left\| \int _{\mathbb {R}^4} Q \cdot c_f(\xi _1,\tau _1) c_g(\xi _2,\tau _2) c_h(\xi -\xi _1-\xi _2,\tau -\tau _1-\tau _2) d\xi _2d\tau _2 d\xi _1d\tau _1\right\| _{L^2}^2\\{} & {} \lesssim \left\| {\varTheta }_1\right\| _{L_{\xi ,\tau }^\infty } \Vert c_f\Vert _{L^2}^2 \Vert c_g\Vert _{L^2}^2\Vert c_h\Vert _{L^2}^2. \end{aligned}$$

Thus, to prove our trilinear estimate (3.5) it suffices to show that the quantity ${\varTheta }_1$ defined below is bounded, that is the following result holds.

Lemma 3.3

If $s> \frac{1}{4}$ and $\frac{1}{2}<b'\le b< \min \{\frac{1}{3}+\frac{2}{3}s,\frac{3}{4}\}$, then for all $\xi$ and $\tau$ we have

$$\begin{aligned} {\varTheta }_1(\xi ,\tau )\doteq & {} \frac{|\xi |^2(1+|\xi |)^{2s}}{(1+|\tau -\xi ^3|)^{2(1-b)}}\nonumber \\{} & {} \times \int _{\mathbb {R}^4}\frac{(1+|\xi _1|)^{-2s} (1+|\xi _2|)^{-2s} (1+|\xi -\xi _1-\xi _2|)^{-2s}\,d\xi _2d\tau _2 d\xi _1d\tau _1}{(1+|\tau _1-\xi _1^3|)^{2b'}(1+|\tau _2-\xi _2^3|)^{2b'}(1+|\tau -\tau _1-\tau _2-(\xi -\xi _1-\xi _2)^3|)^{2b'}}\nonumber \\\lesssim & {} 1. \end{aligned}$$

(3.12)

Proof

We begin the proof by taking advantage of the symmetry in convolution writing of $\widehat{f}*\widehat{g}*\widehat{h}$ to assume the following order of $|\xi _1|$, $|\xi _2|$ and $|\xi -\xi _1-\xi _2|$

$$\begin{aligned} |\xi _1| \ge |\xi _2| \ge |\xi -\xi _1-\xi _2|. \end{aligned}$$

(3.13)

Then, in our $L^2$ formulation of the trilinear estimate (3.5), we can replace the multiplier Q with $\chi _{A}Q$, where A is defined by

$$\begin{aligned} A \doteq \{(\xi ,\xi _1,\xi _2)\in \mathbb {R}^3: |\xi _1| \ge |\xi _2| \ge |\xi -\xi _1-\xi _2|\}. \end{aligned}$$

This results in having $\chi _{A}$ as a factor of the integrand in the quantity ${\varTheta }_1$ defined in (3.12) above. Applying calculus estimate (3.1) in $\tau _1$ with $a=\tau -\tau _2-(\xi -\xi _1-\xi _2)^3$, $c=\xi _1^3$ and $\ell =b'$, from (3.12) we get

$$\begin{aligned} {\varTheta }_1(\xi ,\tau )\lesssim & {} \frac{|\xi |^2(1+|\xi |)^{2s}}{(1+|\tau -\xi ^3|)^{2(1-b)}}\\{} & {} \times \int _{\mathbb {R}^3} \frac{\chi _{A}(\xi ,\xi _1,\xi _2)(1+|\xi _1|)^{-2s}(1+|\xi _2|)^{-2s}(1+|\xi -\xi _1-\xi _2|)^{-2s}}{(1+|\tau _2-\xi _2^3|)^{2b'}(1+|\tau -\tau _2-(\xi -\xi _1-\xi _2)^3-\xi _1^3|)^{2b'}} d\tau _2 d\xi _2 d\xi _1. \end{aligned}$$

Furthermore, applying calculus estimate (3.1) in $\tau _2$ with $a=\tau -(\xi -\xi _1-\xi _2)^3-\xi _1^3$, $c=\xi _2^3$, $\ell =b'$ gives

$$\begin{aligned} {\varTheta }_1(\xi ,\tau )\lesssim & {} \frac{|\xi |^2(1+|\xi |)^{2s}}{(1+|\tau -\xi ^3|)^{2(1-b)}}\\{} & {} \times \int _{\mathbb {R}^2}\frac{\chi _{A}(\xi ,\xi _1,\xi _2)(1+|\xi _1|)^{-2s}(1+|\xi _2|)^{-2s}(1+|\xi -\xi _1-\xi _2|)^{-2s}}{(1+|\tau -\xi ^3-d_3(\xi ,\xi _1,\xi _2)|)^{2b'}}d\xi _2 d\xi _1,\nonumber \end{aligned}$$

(3.14)

where $d_3(\xi ,\xi _1,\xi _2)$ is the Bourgain quantity given by (3.7). Also, since for $(\xi ,\xi _1,\xi _2)\in A$ we have the ordering relation $|\xi _1|\ge |\xi _2|\ge |\xi -\xi _1-\xi _2|$, we obtain the following useful bound for $|\xi |$

$$\begin{aligned} |\xi | = |(\xi -\xi _1-\xi _2)+\xi _2+\xi _1| \le |\xi -\xi _1-\xi _2|+|\xi _2|+|\xi _1| \le 3 |\xi _1|. \end{aligned}$$

(3.15)

Next, we consider the following two cases.

• Case 1: $|\xi _1|\le 100$.

• Case 2: $|\xi _1|> 100$.

Proof in Case 1. Since $|\xi _1|\le 100$ by estimate (3.15) and relation (3.13) we see that all of $|\xi |$, $|\xi _1|$ and $|\xi _2|$ are bounded. Furthermore, since $b'\ge 0$ and $(1-b)\ge 0$, we see that the multiplier and the integrand in (3.14) are bounded. Therefore, ${\varTheta }_1(\xi ,\tau )\lesssim 1$, since the integration is over a bounded set.

Proof in Case 2. We consider the following two subcases.

• Subcase 2.1: $|\xi _1|> 100$ and $|\xi _2|\le \frac{1}{4} |\xi _1|$.

• Subcase 2.2: $|\xi _1|> 100$ and $|\xi _2|> \frac{1}{4} |\xi _1|$.

Proof in Subcase 2.1. Since, by (3.15), $|\xi |\le 3|\xi _1|$ and also $|\xi | = |\xi _1+\xi _2+(\xi -\xi _1-\xi _2)| \ge |\xi _1| - |\xi _2| - |\xi -\xi _1-\xi _2| \ge |\xi _1| - \frac{1}{4}|\xi _1| - \frac{1}{4}|\xi _1| = \frac{1}{2} |\xi _1|$, we have $|\xi | \ge \frac{1}{2} |\xi _1|$, and therefore,

$$\begin{aligned} |\xi |\simeq |\xi _1|. \end{aligned}$$

(3.16)

Using estimate (3.16) we have $(1+|\xi _1|)^{-2s}\simeq (1+|\xi |)^{-2s}$. Also, using $|\xi _2|\ge |\xi -\xi _1-\xi _2|$, for $s\ge 0$ we get $(1+|\xi _2|)^{-2s} (1+|\xi -\xi _1-\xi _2|)^{-2s} \lesssim (1+|\xi -\xi _1-\xi _2|)^{-4s}$. Combining these estimates with (3.14) gives

$$\begin{aligned} {\varTheta }_1(\xi ,\tau ) \lesssim \frac{|\xi |^2}{(1+|\tau -\xi ^3|)^{2(1-b)}} \int _{\mathbb {R}^2} \frac{\chi _{A}(\xi ,\xi _1,\xi _2)(1+|\xi -\xi _1-\xi _2|)^{-4s}}{(1+|\tau -\xi ^3 - d_3(\xi ,\xi _1,\xi _2)|)^{2b'}} d\xi _2 d\xi _1. \end{aligned}$$

(3.17)

Now, making the change of variables $\widetilde{\xi }_2=\xi -\xi _1-\xi _2$ (or $\xi _2=\xi -\xi _1-\widetilde{\xi }_2$) and $\widetilde{\xi }_1=\xi _1$, and using the relation $d_3(\xi ,\xi _1,\xi _2) = -\xi ^3+\xi _1^3+\xi _2^3+(\xi - \xi _1-\xi _2)^3 = d_3(\xi ,\widetilde{\xi }_1,\widetilde{\xi }_2)$, from (3.17) we get

$$\begin{aligned} {\varTheta }_1(\xi ,\tau ) \lesssim \frac{|\xi |^2}{(1+|\tau -\xi ^3|)^{2(1-b)}} \int _{\mathbb {R}^2} \frac{\chi _{\widetilde{A}}(\xi ,\widetilde{\xi }_1,\widetilde{\xi }_2) (1+|\widetilde{\xi }_2|)^{-4s}}{(1+|\tau -\xi ^3 - d_3(\xi ,\widetilde{\xi }_1,\widetilde{\xi }_2)|)^{2b'}} d\widetilde{\xi }_2 d\widetilde{\xi }_1, \end{aligned}$$

(3.18)

where the domain $\widetilde{A}$ is given by $\widetilde{A}\doteq \{(\xi ,\widetilde{\xi }_1,\widetilde{\xi }_2)\in \mathbb {R}^3 : |\widetilde{\xi }_1| \ge |\xi -\widetilde{\xi }_1-\widetilde{\xi }_2| \ge |\widetilde{\xi }_2|\}$. Furthermore, using Fubini’s theorem to switch the integration of $d\widetilde{\xi }_2$ and $d\widetilde{\xi }_1$, from (3.18) we get

$$\begin{aligned} \begin{aligned} {\varTheta }_1(\xi ,\tau )&\lesssim \frac{|\xi |^2}{(1+|\tau -\xi ^3|)^{2(1-b)}}\\&\quad \times \int _{\mathbb {R}} (1+|\widetilde{\xi }_2|)^{-4s}\left[ \int _{\mathbb {R}}\frac{\chi _{\widetilde{A}}(\xi ,\widetilde{\xi }_1,\widetilde{\xi }_2)}{(1+|\tau -\xi ^3 - d_3(\xi ,\widetilde{\xi }_1,\widetilde{\xi }_2) |)^{2b'}} d\widetilde{\xi }_1\right] d\widetilde{\xi }_2. \end{aligned} \end{aligned}$$

(3.19)

For the $d\widetilde{\xi }_1$-integral in (3.19) our strategy is to make the change of variables $\mu =\mu (\widetilde{\xi }_1)=d_3(\xi ,\widetilde{\xi }_1,\widetilde{\xi }_2)$. For this change to be good, we need to split the $\widetilde{\xi }_1$-integral at the critical point of $\mu (\widetilde{\xi }_1)$, which is $p=(\xi -\widetilde{\xi }_2)/2$, since $\mu '(\widetilde{\xi }_1) =\partial d_3(\xi ,\widetilde{\xi }_1,\widetilde{\xi }_2)/\partial \widetilde{\xi }_1 = 6(\xi -\widetilde{\xi }_2)\big [\widetilde{\xi }_1-\frac{1}{2}(\xi -\widetilde{\xi }_2)\big ]$ (see property (3.9)). Thus, using $\widetilde{\xi }_1$ intervals $I_1^{\widetilde{\xi }_1}\doteq \big (-\infty , (\xi -\widetilde{\xi }_2)/2 \big )$ and $I_2^{\widetilde{\xi }_1}\doteq \big ( (\xi -\widetilde{\xi }_2)/2,\infty \big )$, making the change of variables $\mu =\mu (\widetilde{\xi }_1)=d_3(\xi ,\widetilde{\xi }_1,\widetilde{\xi }_2)$ on each one of these two intervals and defining $I_k^\mu$ to be the range of $\mu$ when $\widetilde{\xi }_1\in I_k^{\widetilde{\xi }_1}$, from (3.19) we get

$$\begin{aligned} {\varTheta }_1 \lesssim J_1+J_2, \end{aligned}$$

(3.20)

with

$$\begin{aligned} J_k \doteq \frac{|\xi |^2}{(1+|\tau -\xi ^3|)^{2(1-b)}} \int _{\mathbb {R}} (1+|\widetilde{\xi }_2|)^{-4s} \left( \int _{I_k^\mu } \frac{d\mu }{(1+|\tau -\xi ^3-\mu |)^{2b'} |\mu '|}\right) d\widetilde{\xi }_2. \end{aligned}$$

Now, we need the following bound of $|\mu '|$ from below

$$\begin{aligned} |\mu '| = 6|\xi -\widetilde{\xi }_2| \left| \widetilde{\xi }_1-(\xi -\widetilde{\xi }_2)/2\right| \gtrsim |\xi |^2, \end{aligned}$$

(3.21)

which follows from $|\xi -\widetilde{\xi }_2| \ge |\widetilde{\xi }_1| - |\xi -\widetilde{\xi }_1-\widetilde{\xi }_2| \ge |\xi _1|-\frac{1}{4}|\xi _1| = \frac{3}{4}|\xi _1| \gtrsim |\xi |$ and $|\widetilde{\xi }_1-\frac{1}{2}(\xi -\widetilde{\xi }_2)| = |\frac{1}{2}\widetilde{\xi }_1-\frac{1}{2}(\xi -\widetilde{\xi }_1-\widetilde{\xi }_2)| \ge \frac{1}{2}|\widetilde{\xi }_1| - \frac{1}{2}|\xi -\widetilde{\xi }_1-\widetilde{\xi }_2| \ge \frac{1}{2}|\xi _1|-\frac{1}{8}|\xi _1| = \frac{3}{8}|\xi _1| \gtrsim |\xi |$, since $\widetilde{\xi }_1=\xi _1$ and $|\xi -\widetilde{\xi }_1-\widetilde{\xi }_2|= |\xi _2|\le \frac{1}{4}|\xi _1|$. Next, combining estimate (3.21) with (3.20) and integrating $\mu$ over $\mathbb {R}$, we get

$$\begin{aligned} J_k \lesssim \frac{1}{(1+|\tau -\xi ^3|)^{2(1-b)}} \int _{\mathbb {R}} \frac{d\widetilde{\xi }_2}{(1+|\widetilde{\xi }_2|)^{4s}} \cdot \int _{\mathbb {R}} \frac{d\mu }{(1+|\tau -\xi ^3- \mu |)^{2b'}}, \quad k=1,2. \end{aligned}$$

The multiplier $\frac{1}{(1+|\tau -\xi ^3|)^{2(1-b)}}$ is bounded if $2(1-b)\ge 0$ or $b\le 1$. Also, the first integral is bounded if $4s>1$ or $s>\frac{1}{4}$, and the second integral is bounded if $2b'>1$ or $b'>\frac{1}{2}$. Therefore, for $J_k$ to be bounded it suffices to have $\frac{1}{2}<b'\le b\le 1$ and $s>\frac{1}{4}$. This completes the proof in Subcase 2.1.

Proof in Subcase 2.2. Since $\frac{1}{2}<b'\le b<1$ we have $0<2(1-b)<1<2b'$, so we can move $(1+|\tau -\xi ^3|)^{2(1-b)}$ inside the integral and replace $2b'$ with $2(1-b)$. Since $(1+|\tau -\xi ^3|)(1+|\tau -\xi ^3-d_3(\xi ,\xi _1,\xi _2)|) \ge |\tau -\xi ^3|+|\tau -\xi ^3-d_3(\xi ,\xi _1,\xi _2)|$ and also $|\tau -\xi ^3|+|\tau -\xi ^3-d_3(\xi ,\xi _1,\xi _2)| \ge |\tau -\xi ^3-d_3(\xi ,\xi _1,\xi _2) - (\tau -\xi ^3)| = |d_3(\xi ,\xi _1,\xi _2)|$, we have

$$\begin{aligned} (1+|\tau -\xi ^3|)^{2(1-b)} (1+ |\tau -\xi ^3-d_3(\xi ,\xi _1,\xi _2)|)^{2b'} \gtrsim |d_3(\xi ,\xi _1,\xi _2)|^{2(1-b)}. \end{aligned}$$

(3.22)

Using (3.22), from (3.14) we obtain

$$\begin{aligned} {\varTheta }_1(\xi ,\tau )\lesssim & {} |\xi |^2(1+|\xi |)^{2s}\\{} & {} \times \int _{\mathbb {R}^2} \frac{\chi _{A}(\xi ,\xi _1,\xi _2)(1+|\xi _1|)^{-2s} (1+|\xi _2|)^{-2s} (1+|\xi -\xi _1-\xi _2|)^{-2s}}{|d_3(\xi ,\xi _1,\xi _2)|^{2(1-b)}} d\xi _2 d\xi _1.\nonumber \end{aligned}$$

(3.23)

Recalling the factorization $d_3=-3(\xi -\xi _1)(\xi -\xi _2)(\xi _1+\xi _2)$, from (3.23) we obtain

$$\begin{aligned} {\varTheta }_1(\xi ,\tau )\lesssim & {} |\xi |^2(1+|\xi |)^{2s}\\{} & {} \times \int _{\mathbb {R}^2} \frac{\chi _{A}(\xi ,\xi _1,\xi _2)(1+|\xi _1|)^{-2s}(1+|\xi _2|)^{-2s}(1+|\xi -\xi _1-\xi _2|)^{-2s}}{|(\xi -\xi _1)(\xi -\xi _2)(\xi _1+\xi _2)|^{2(1-b)}} d\xi _2 d\xi _1.\nonumber \end{aligned}$$

(3.24)

Next, we consider the following two possibilities about the size of $|\xi -\xi _1-\xi _2|$.

• Subcase 2.2.1: $|\xi _1|> 100$, $|\xi _2|> \frac{1}{4} |\xi _1|$ and $|\xi -\xi _1-\xi _2|\le \frac{1}{16}|\xi _1|$.

• Subcase 2.2.2: $|\xi _1|> 100$, $|\xi _2|> \frac{1}{4} |\xi _1|$ and $|\xi -\xi _1-\xi _2|> \frac{1}{16}|\xi _1|$.

Proof in Subcase 2.2.1. In this situation we have $|\xi -\xi _1|=|\xi _2+(\xi -\xi _1-\xi _2)|\ge |\xi _2|-|\xi -\xi _1-\xi _2|\ge (\frac{1}{4}-\frac{1}{16})|\xi _1|\gtrsim |\xi _1|$ and $|\xi -\xi _2|=|\xi _1+(\xi -\xi _1-\xi _2)|\ge |\xi _1|-|\xi -\xi _1-\xi _2|\ge (1-\frac{1}{16})|\xi _1|\gtrsim |\xi _1|$. Thus,

$$\begin{aligned} |\xi -\xi _1| \gtrsim |\xi _1| \quad \text { and } \quad |\xi -\xi _2| \gtrsim |\xi _1|. \end{aligned}$$

(3.25)

Now, using estimates (3.25), $(1+|\xi _1|)^{-2s}\simeq |\xi _1|^{-2s}$ and $(1+|\xi _2|)^{-2s}\lesssim |\xi _1|^{-2s}$, putting $|\xi |^2(1+|\xi |)^{2s}$ inside the integral, from (3.24) we obtain

$$\begin{aligned} {\varTheta }_1(\xi ,\tau ) \lesssim \int _{\mathbb {R}} \frac{|\xi |^2(1+|\xi |)^{2s}}{|\xi _1|^{4-4b+4s}}\left[ \int _{\mathbb {R}} \frac{\chi _{A}(\xi ,\xi _1,\xi _2)(1+|\xi -\xi _1-\xi _2|)^{-2s}}{|\xi _1+\xi _2|^{2(1-b)}}d\xi _2\right] d\xi _1. \end{aligned}$$

(3.26)

Furthermore, we split the $\xi _2$-integral in (3.26) for $|\xi _1+\xi _2|\le 1$ and for $|\xi _1+\xi _2|>1$ by using the $\xi _2$ intervals $I_1^{\xi _2}=(-1-\xi _1,1-\xi _1)$ and $I_2^{\xi _2}=(-\infty ,-1-\xi _1)\cup (1-\xi _1,\infty )$. Thus, from (3.26) we have

$$\begin{aligned} {\varTheta }_1 \lesssim J_1+J_2 \end{aligned}$$

(3.27)

with

$$ J_k \doteq \int _{\mathbb {R}} \frac{|\xi |^2(1+|\xi |)^{2s}}{|\xi _1|^{4-4b+4s}} \left[ \int _{I_k^{\xi _2}} \frac{\chi _{A}(\xi ,\xi _1,\xi _2)(1+|\xi -\xi _1-\xi _2|)^{-2s}}{|\xi _1+\xi _2|^{2(1-b)}}d\xi _2\right] d\xi _1. $$

Estimate for $J_1$. Since $\xi _2\in I_1^{\xi _2}$ we have $|\xi _1+\xi _2|\lesssim 1$, which implies that $1+|\xi | \simeq 1+|\xi -(\xi _1+\xi _2)|$. This gives $(1+|\xi -\xi _1-\xi _2|)^{-2s}\simeq (1+|\xi |)^{-2s}$. Thus, from (3.27) we obtain

$$\begin{aligned} J_1 \lesssim \int _{\mathbb {R}} \frac{|\xi |^2}{|\xi _1|^{4-4b+4s}} \left[ \int _{I_1^{\xi _2}} \frac{\chi _{A}(\xi ,\xi _1,\xi _2)}{|\xi _1+\xi _2|^{2(1-b)}} d\xi _2\right] d\xi _1. \end{aligned}$$

(3.28)

Now, making the change of variables $\mu =\mu (\xi _2)=\xi _1+\xi _2$ and using $|\mu |\le 1$, from (3.28) we obtain

$$\begin{aligned} J_1 \lesssim \int _{|\mu |\le 1} \frac{1}{|\mu |^{2(1-b)}}d\mu \cdot \int _{\mathbb {R}} \frac{|\xi |^2}{|\xi _1|^{4-4b+4s}} d\xi _1. \end{aligned}$$

(3.29)

For the first integral in the above estimate, since $2(1-b)<1$ or $b>\frac{1}{2}$, we have $\int _{|\mu |\le 1}\frac{1}{|\mu |^{2(1-b)}}d\mu \simeq (\mu ^{2b-1})\big |_0^1 = 1$. For the second integral in (3.29), using estimate (3.15), i.e. $|\xi |\le 3|\xi _1|$, we get $\int _{\mathbb {R}} \frac{|\xi |^2}{|\xi _1|^{4-4b+4s}}d\xi _1 \lesssim \int _{\mathbb {R}}\frac{1}{|\xi _1|^{2-4b+4s}}d\xi _1$, which is bounded if $2-4b+4s>1$ or $b<\frac{1}{4}+s$, since $|\xi _1|>100$. For $b>\frac{1}{2}$, it suffices to have $\frac{1}{4}+s>\frac{1}{2}$ or $s>\frac{1}{4}$. Combining the above computations with (3.29), for $J_1$ to be bounded, it suffices to have $\frac{1}{2}<b<\frac{1}{4}+s$ and $s>\frac{1}{4}$. This completes the proof of boundedness for $J_1$.

Estimate for $J_2$. Since $\xi _2\in I_2^{\xi _2}$, we have $|\xi _1+\xi _2|>1$, which implies that $|\xi _1+\xi _2|\simeq (1+|\xi _1+\xi _2|)$ or $|\xi _1+\xi _2|^{2(1-b)}\simeq (1+|\xi _1+\xi _2|)^{2(1-b)}$. Using this, from (3.27) we get

$$\begin{aligned} J_2 \lesssim \int _{\mathbb {R}} \frac{|\xi |^2(1+|\xi |)^{2s}}{|\xi _1|^{4-4b+4s}}\left[ \int _{I_2^{\xi _2}} \frac{\chi _{A}(\xi ,\xi _1,\xi _2)}{(1+|\xi -\xi _1-\xi _2|)^{2s}(1+|\xi _1+\xi _2|)^{2(1-b)}}d\xi _2\right] d\xi _1. \end{aligned}$$

(3.30)

Now, for the $\xi _2$-integral in (3.30), making the change of variables $\mu =\mu (\xi _2)=\xi _1+\xi _2$ and using $|\mu |\le |\xi _1|+|\xi _2|\le 2|\xi _1|$, we get

$$\begin{aligned} J_2 \lesssim \int _{\mathbb {R}} \frac{|\xi |^2 (1+|\xi |)^{2s}}{|\xi _1|^{4-4b+4s}}\left[ \int _{|\mu |\le 2|\xi _1|}\frac{1}{(1+|\xi -\mu |)^{2s}(1+|\mu |)^{2(1-b)}} d\mu \right] d\xi _1. \end{aligned}$$

(3.31)

Next, we consider the following two cases concerning s.

• Subcase 2.2.1.1: $\frac{1}{4}<s<\frac{1}{2}$.

• Subcase 2.2.1.2: $s\ge \frac{1}{2}$.

Proof in Subcase 2.2.1.1. For the $d\mu$-integral in (3.31), integrating it over $\mathbb {R}$ and using calculus estimate (3.3) with $\frac{1}{4}<\ell =s<\frac{1}{2}$ and $\frac{1}{4}<\ell '=1-b<\frac{1}{2}$ (or $\frac{1}{2}<b<\frac{3}{4}$), $a=\xi$ and $c=0$, we get

$$\begin{aligned} \int _{|\mu |\le 2|\xi _1|}\frac{1}{(1+|\xi -\mu |)^{2s}(1+|\mu |)^{2(1-b)}}d\mu\le & {} \int _{\mathbb {R}} \frac{1}{(1+|\xi -\mu |)^{2s}(1+|\mu |)^{2(1-b)}}d\mu \\\lesssim & {} \frac{1}{(1+|\xi |)^{2s+1-2b}}. \end{aligned}$$

Combining the above estimate with (3.31), we obtain

$$\begin{aligned} J_2 \lesssim \int _{\mathbb {R}} \frac{|\xi |^2 (1+|\xi |)^{2s}}{|\xi _1|^{4-4b+4s}} \frac{1}{(1+|\xi |)^{2s+1-2b}} d\xi _1 = \int _{\mathbb {R}}\frac{|\xi |^2(1+|\xi |)^{2b-1}}{|\xi _1|^{4-4b+4s}}d\xi _1. \end{aligned}$$

(3.32)

Again, using estimate (3.15), i.e. $|\xi |\le 3|\xi _1|$, and $2b-1>0$ from (3.32) we get $J_2\lesssim \int _{\mathbb {R}}\frac{1}{|\xi _1|^{3-6b+4s}}d\xi _1$, which is bounded if $3-6b+4s>1$ or $b<\frac{2}{3}s+\frac{1}{3}$. For $b>\frac{1}{2}$, it suffices to have $\frac{2}{3}s+\frac{1}{3}>\frac{1}{2}$ or $\frac{2}{3}s>\frac{1}{6}$, which implies that $s>\frac{1}{4}$. This completes the estimate for $J_2$ when $\frac{1}{4}<s<\frac{1}{2}$.

Proof in Subcase 2.2.1.2 ($s\ge \frac{1}{2}$). Using $\mu$-intervals $I_1^{\mu }= \{\mu \in \mathbb {R}: |\mu |\le 2|\xi _1| \text { and }|\xi -\mu |\le \frac{1}{2} |\xi | \}$, $I_2^{\mu }=\{\mu \in \mathbb {R}: |\mu |\le 2|\xi _1| \text { and }|\xi -\mu |>\frac{1}{2}|\xi |\}$, we split $J_2$ as $J_2 \lesssim J_{21} + J_{22}$, where

$$\begin{aligned} J_{2k} \doteq \int _{\mathbb {R}} \frac{|\xi |^2(1+|\xi |)^{2s}}{|\xi _1|^{4-4b+4s}}\left[ \int _{I_k^\mu }\frac{d\mu }{(1+|\xi -\mu |)^{2s}(1+|\mu |)^{2(1-b)}}\right] d\xi _1,\quad k=1,2. \end{aligned}$$

(3.33)

Estimate for $J_{21}$. For $\mu \in I_1^\mu$, we have

$$\begin{aligned} |\mu | \gtrsim |\xi |, \end{aligned}$$

(3.34)

which follows from $|\mu | = |\xi -(\xi -\mu )| \ge |\xi |-|\xi -\mu | \ge |\xi |-\frac{1}{2}|\xi | = \frac{1}{2}|\xi |$, since $|\xi -\mu |\le \frac{1}{2}|\xi |$. Also, since $2(1-b)\ge 0$ or $b\le 1$, using estimate (3.34) we get $\frac{1}{(1+|\mu |)^{2(1-b)}} \lesssim \frac{1}{(1+|\xi |)^{2(1-b)}}$, which combined with estimate (3.33) gives

$$\begin{aligned} J_{21} \lesssim \int _{\mathbb {R}} \frac{|\xi |^2 (1+|\xi |)^{2s-2(1-b)}}{|\xi _1|^{4-4b+4s}}\left[ \int _{I_1^\mu } \frac{d\mu }{(1+|\xi -\mu |)^{2s}}\right] d\xi _1. \end{aligned}$$

(3.35)

For the $d\mu$-integral in (3.35), we make the change of variables $\mu _1=\xi -\mu$. Using estimate (3.15) and $|\mu |\le 2|\xi _1|$, we get $|\mu _1|\le |\xi |+|\mu |\le 5|\xi _1|$. Also, since $2s\ge 1$ we obtain $\int _{I_1^\mu }\frac{d\mu }{(1+|\xi -\mu |)^{2s}} \lesssim \int _{|\mu _1|\lesssim |\xi _1|} \frac{d\mu _1}{(1+|\mu _1|)^{2s}}\lesssim \ln |\xi _1|$. Again, using estimate (3.15) we get $|\xi |^2 (1+|\xi |)^{2s-2(1-b)}\lesssim (1+|\xi |)^{2s+2b}\lesssim |\xi _1|^{2s+2b}$. Combining the above computations with (3.35), we obtain $J_{21}\lesssim \int _{\mathbb {R}} \frac{|\xi _1|^{2s+2b}}{|\xi _1| ^{4-4b+4s}} \cdot \ln |\xi _1| d\xi _1 = \int _{\mathbb {R}} \frac{1}{|\xi _1|^{4-6b+2s}} \cdot \ln |\xi _1| d\xi _1$, which is bounded if $4-6b+2s>1$ or $b<\frac{1}{2}+\frac{s}{3}$. For $b>\frac{1}{2}$, it suffices to have $s>0$. This completes the estimate for $J_{21}$.

Estimate for $J_{22}$. Since $|\xi -\mu |\ge \frac{1}{2}|\xi |$, for $s\ge 0$ we have $\frac{1}{(1+|\xi -\mu |)^{2s}}\lesssim \frac{1}{(1+|\xi |)^{2s}}$ or $\frac{(1+|\xi |)^{2s}}{(1+|\xi -\mu |)^{2s}}\lesssim 1$. Putting $(1+|\xi |)^{2s}$ inside the $d\mu$ integral, from (3.33) we get

$$\begin{aligned} J_{22} \lesssim \int _{\mathbb {R}} \frac{|\xi |^2}{|\xi _1|^{4-4b+4s}} \left[ \int _{I_2^\mu } \frac{d\mu }{(1+|\mu |)^{2(1-b)}}\right] d\xi _1. \end{aligned}$$

(3.36)

For the $d\mu$-integral in the above estimate, since $2(1-b)<1$ or $b>\frac{1}{2}$ and $|\mu |\le 2|\xi _1|$ we have $\int _{I_2^\mu }\frac{d\mu }{(1+|\mu |)^{2(1-b)}} \simeq [(1+\mu )^{2b-1}]\big |_0^{2|\xi _1|}\simeq |\xi _1|^{2b-1}$. Again, using estimate (3.15) and the above computation, from (3.36) we obtain

$$\begin{aligned} J_{22} \lesssim \int _{\mathbb {R}} \frac{|\xi _1|^2}{(1+|\xi _1|)^{4-4b+4s}} |\xi _1|^{2b-1} d\xi _1 = \int _{\mathbb {R}} \frac{1}{(1+|\xi _1|)^{3-6b+4s}} d\xi _1, \end{aligned}$$

which is bounded if $3-6b+4s>1$ or $b<\frac{1}{3}+\frac{2}{3}s$. For $b>\frac{1}{2}$, it suffices to have $\frac{1}{3}+\frac{2}{3}s>\frac{1}{2}$ or $\frac{2}{3}s>\frac{1}{6}$, which implies that $s>\frac{1}{4}$. This completes the proof of Subcase 2.2.1.

Proof in Subcase 2.2.2. Since $|\xi _1|> 100$, $|\xi _2|> \frac{1}{4} |\xi _1|$ and $|\xi -\xi _1-\xi _2|> \frac{1}{16}|\xi _1|$, we have the relations

$$\begin{aligned} |\xi -\xi _1-\xi _2| \simeq |\xi _2| \simeq |\xi _1| > 100. \end{aligned}$$

(3.37)

Now, moving $|\xi |^2(1+|\xi |)^{2s}$ inside the integral in (3.24), we get

$$\begin{aligned}{} & {} {\varTheta }_1(\xi ,\tau )\\\lesssim & {} \int _{\mathbb {R}^2} \frac{\chi _{A}(\xi ,\xi _1,\xi _2)|\xi |^2(1+|\xi |)^{2s} (1+|\xi _1|)^{-2s} (1+|\xi _2|)^{-2s}(1+|\xi -\xi _1-\xi _2|)^{-2s}}{|(\xi -\xi _1)(\xi -\xi _2)(\xi _1+\xi _2)|^{2(1-b)}}d\xi _2 d\xi _1 \end{aligned}$$

and simplifying the numerator as follows, by using estimate (3.15), i.e. $|\xi |\le 3|\xi _1|$, and (3.37),

$$ |\xi |^2(1+|\xi |)^{2s}(1+|\xi _1|)^{-2s} (1+|\xi _2|)^{-2s} (1+|\xi -\xi _1-\xi _2|)^{-2s} \lesssim |\xi _1|^{1-2s}|\xi _2|^{1-2s}, $$

we are reduced to the inequality

$$\begin{aligned} {\varTheta }_1(\xi ,\tau ) \lesssim \int _{\mathbb {R}^2} \frac{\chi _{A}(\xi ,\xi _1,\xi _2) |\xi _1|^{1-2s} |\xi _2|^{1-2s}}{|(\xi -\xi _1)(\xi -\xi _2)(\xi _1+\xi _2) |^{2(1-b)}}d\xi _2 d\xi _1. \end{aligned}$$

(3.38)

Next, we consider the following two possibilities.

• Subcase 2.2.2.1: $|\xi _1|> 100$, $|\xi _2|> \frac{1}{4} |\xi _1|$, $|\xi -\xi _1-\xi _2|> \frac{1}{16}|\xi _1|$ and $\big (|\xi |\le \frac{1}{64}|\xi -\xi _1-\xi _2|$ or $|\xi |\ge \frac{65}{64}|\xi _1|\big )$.

• Subcase 2.2.2.2: $|\xi _1|> 100$, $|\xi _2|> \frac{1}{4} |\xi _1|$, $|\xi -\xi _1-\xi _2|> \frac{1}{16}|\xi _1|$ and $\frac{1}{64}|\xi -\xi _1-\xi _2|< |\xi |< \frac{65}{64}|\xi _1|$.

Proof in Subcase 2.2.2.1. For each factor of the denominator in (3.38) we have the following estimate

$$\begin{aligned} |\xi -\xi _1|\ge & {} \big ||\xi |-|\xi _1|\big | \ge \frac{1}{64}|\xi _1| \gtrsim |\xi _1| \simeq |\xi _2|, \end{aligned}$$

(3.39a)

$$\begin{aligned} |\xi -\xi _2|\ge & {} \big ||\xi |-|\xi _2|\big | \ge \frac{1}{64}|\xi _2| \gtrsim |\xi _2| \simeq |\xi _1|, \end{aligned}$$

(3.39b)

$$\begin{aligned} |\xi _1+\xi _2|= & {} |\xi -(\xi -\xi _1-\xi _2)| \ge \big ||\xi |-|\xi -\xi _1-\xi _2|\big |\nonumber \\\ge & {} \frac{1}{64}|\xi -\xi _1-\xi _2| \simeq |\xi _1| \simeq |\xi _2|. \end{aligned}$$

(3.39c)

This follows from estimate (3.37) and $|\xi |\le \frac{1}{64}|\xi -\xi _1-\xi _2|\le \frac{1}{64}|\xi _2|\le \frac{1}{64}|\xi _1|$ or $|\xi |\ge \frac{65}{64}|\xi _1|\ge \frac{65}{64}|\xi _2|\ge \frac{65}{64}|\xi -\xi _1-\xi _2|$. Using estimates (3.39a)–(3.39c), we get $|(\xi -\xi _1)(\xi -\xi _2)(\xi _1+\xi _2)|\gtrsim |\xi _1|^{\frac{3}{2}}|\xi _2|^{\frac{3}{2}}$. Using this and estimate (3.37), from (3.38) we obtain

$$\begin{aligned} {\varTheta }_1(\xi ,\tau )\lesssim & {} \int _{|\xi _2|\simeq |\xi _1|> 100} \frac{|\xi _1|^{1-2s}|\xi _2|^{1-2s}}{|\xi _1|^{3(1-b)}|\xi _2|^{3(1-b)}}d\xi _2 d\xi _1\\= & {} \int _{|\xi _1|\gtrsim 1} \frac{1}{|\xi _1|^{3(1-b)-1+2s}} d\xi _1 \int _{|\xi _2|\gtrsim 1} \frac{1}{|\xi _2|^{3(1-b)-1+2s}} d\xi _2, \end{aligned}$$

which is bounded if $3(1-b)-1+2s>1$ or $3(1-b) > 2-2s$. It suffices to have $1-b > \frac{2}{3}-\frac{2}{3}s$ or $b<\frac{1}{3}+\frac{2}{3}s$. For $b>\frac{1}{2}$, it suffices to have $\frac{1}{3}+\frac{2}{3}s>\frac{1}{2}$ or $\frac{2}{3}s>\frac{1}{6}$, which implies $s>\frac{1}{4}$. This completes the proof in Subcase 2.2.2.1.

Proof in Subcase 2.2.2.2. Using estimate (3.37) and $\frac{1}{64}|\xi -\xi _1-\xi _2|< |\xi |< \frac{65}{64}|\xi _1|$, we have

$$\begin{aligned} |\xi | \simeq |\xi -\xi _1-\xi _2|\simeq |\xi _2|\simeq |\xi _1| > 100. \end{aligned}$$

(3.40)

Using (3.40) we bound the numerator in (3.24) as follows $|\xi _1|^{1-2s}|\xi _2|^{1-2s} \simeq |\xi |^{2-4s}$. Taking out $|\xi |^{2-4s}$, from (3.38) we get

$$\begin{aligned} {\varTheta }_1(\xi ,\tau ) \lesssim |\xi |^{2-4s} \int _{\mathbb {R}^2} \frac{1}{|(\xi -\xi _1)(\xi -\xi _2)(\xi _1+\xi _2)|^{2(1-b)}}d\xi _1 d\xi _2. \end{aligned}$$

(3.41)

Now, it suffices to estimate (3.41) in the case we integrate over $E_1$ (the other two cases are similar)

$$\begin{aligned} E_1 = \{(\xi _1,\xi _2) \in \mathbb {R}^2: |\xi -\xi _1| \ge \max \{ |\xi -\xi _2|,|\xi _1+\xi _2|\}\}. \end{aligned}$$

(3.42)

Using the assumption (3.42) and the triangle inequality we get

$$\begin{aligned} \begin{aligned} |\xi -\xi _1|&\ge \frac{1}{3}\left[ |\xi -\xi _1| + |\xi -\xi _2| + |\xi _1+\xi _2|\right] \\&\ge \frac{1}{3} |(\xi -\xi _1)+(\xi -\xi _2)+(\xi _1+\xi _2)| = \frac{2}{3}|\xi | \gtrsim |\xi |. \end{aligned} \end{aligned}$$

(3.43)

Now, using estimate (3.43), from (3.41) we get

$$\begin{aligned} \begin{aligned} {\varTheta }_1&\lesssim |\xi |^{2-4s} \int _{E_1} \frac{d\xi _1 d\xi _2}{|\xi |^{2(1-b)} |(\xi -\xi _2)(\xi _1+\xi _2)|^{2(1-b)}}\\&= |\xi |^{2b-4s} \int _{E_1} \frac{d\xi _1d\xi _2}{|(\xi -\xi _2)(\xi _1+\xi _2)|^{2(1-b)}}. \end{aligned} \end{aligned}$$

(3.44)

Furthermore, making the change of variables $\mu _1=\xi -\xi _2$, $\mu _2=\xi _1+\xi _2$ and using estimate (3.40), we get $|\mu _1|=|\xi -\xi _2|\le |\xi |+|\xi _2|\lesssim |\xi |$, $|\mu _2|=|\xi _1+\xi _2|\le |\xi _1|+|\xi _2|\lesssim |\xi |$. Thus, from (3.44) we obtain

$$\begin{aligned} {\varTheta }_1(\xi ,\tau ) \lesssim |\xi |^{2b-4s} \int _{|\mu _1|\lesssim |\xi |} \frac{1}{|\mu _1|^{2(1-b)}} d\mu _1 \int _{|\mu _2|\lesssim |\xi |}\frac{1}{|\mu _2|^{2(1-b)}}d\mu _2. \end{aligned}$$

(3.45)

For the first integral in the above estimate, since $2(1-b)<1$, we have $\int _{|\mu _1|\lesssim |\xi |}\frac{1}{|\mu _1|^{2(1-b)}}$ $d\mu _1\simeq (\mu _1^{2b-1})|_0^{|\xi |}=|\xi |^{2b-1}$. Similarly, we have $\int _{|\mu _2|\lesssim |\xi |}\frac{1}{|\mu _2|^{2(1-b)}}d\mu _1\simeq |\xi |^{2b-1}$. Using these computations, from (3.45) we get

$$\begin{aligned} {\varTheta }_1(\xi ,\tau ) \lesssim |\xi |^{2b-4s} \cdot |\xi |^{2b-1} \cdot |\xi |^{2b-1} = |\xi |^{6b-4s-2}. \end{aligned}$$

Using (3.40), we get $|\xi |\gtrsim 1$. Thus, the above quantity is bounded if $6b-4s-2\le 0$ or $b\le \frac{1}{3}+\frac{2}{3}s$. For $b>\frac{1}{2}$ it suffices to have $\frac{1}{3}+\frac{2}{3}s > \frac{1}{2}$ or $\frac{2}{3}s>\frac{1}{6}$, which gives $s>1/4$. This completes the proof of Lemma 3.3, and also the proof of the mKdV trilinear estimate (2.16). $\square$

3.2 Proof of the First Nonlocal Trilinear Estimates (2.17)

We recall that this estimate reads as follows: $\Vert (1+\partial _x^2)^{-1}\partial _x[(\partial _xf)(\partial _xg)h]\Vert _{X^{s,b-1}}\le$ $c_{s,b}\Vert f\Vert _{X^{s,b'}}\Vert g\Vert _{X^{s,b'}}\Vert h\Vert _{X^{s,b'}}$. Since $(\partial _xf)\cdot (\partial _xg)\cdot h$ is not symmetric, which can be seen clearly if we write it in its $L^2$ form using convolution, we will symmetrize it by introducing the following symmetric quantity

$$\begin{aligned} \begin{aligned} \mathcal {T}(f,g,h)&\doteq \Vert (1+\partial _x^2)^{-1}\partial _x[(\partial _xf)(\partial _xg)h]\Vert _{X^{s,b-1}} + \Vert (1+\partial _x^2)^{-1}\partial _x[(\partial _xg)(\partial _xh)f]\Vert _{X^{s,b-1}}\\&\quad + \Vert (1+\partial _x^2)^{-1}\partial _x[(\partial _xh)(\partial _xf)g]\Vert _{X^{s,b-1}}\\&\le c_{s,b}\Vert f\Vert _{X^{s,b'}}\Vert g\Vert _{X^{s,b'}}\Vert h\Vert _{X^{s,b'}}. \end{aligned} \end{aligned}$$

(3.46)

Then proving the trilinear estimate for $\mathcal {T}(f,g,h)$ gives the desired nonlocal trilinear estimates (2.17). Using $a^2+b^2+c^2\le (|a|+|b|+|c|)^2\le 3(a^2+b^2+c^2)$ we bound $\mathcal {T}(f,g,h)$ as follows

$$\begin{aligned} |\mathcal {T}(f,g,h)|^2\lesssim & {} \int _{\mathbb {R}^2} \frac{|\xi |^2(1+|\xi |)^{2s}}{(1+\xi ^2)^2(1+|\tau -\xi ^3|)^{2(1-b)}}\\{} & {} \times \bigg [\int _{\mathbb {R}^4} (|\xi _1||\xi _2|+|\xi _2||\xi -\xi _1-\xi _2|+|\xi -\xi _1-\xi _2||\xi _1|)\\{} & {} |\widehat{f}(\xi _1,\tau _1)\widehat{g}(\xi _2,\tau _2)\widehat{h}(\xi -\xi _1-\xi _2,\tau -\tau _1-\tau _2)|d\xi _2d\tau _2d\xi _1d\tau _1\bigg ]^2d\xi d\tau . \end{aligned}$$

Furthermore, using notation (3.4) we see that estimate (3.46) follows from its $L^2$ formulation

$$\begin{aligned} \begin{aligned}&\left\| \int _{\mathbb {R}^4} Q\cdot c_f(\xi _1,\tau _1)c_g(\xi _2,\tau _2) c_h(\xi -\xi _1-\xi _2,\tau -\tau _1-\tau _2) d\xi _2d\tau _2 d\xi _1d\tau _1\right\| _{L^2_{\xi ,\tau }}\\&\lesssim \Vert c_f\Vert _{L^2_{\xi ,\tau }}\Vert c_g\Vert _{L^2_{\xi ,\tau }}\Vert c_h\Vert _{L^2_{\xi ,\tau }}, \end{aligned} \end{aligned}$$

(3.47)

where the multiplier $Q=Q(\xi ,\tau ,\xi _1,\tau _1,\xi _2,\tau _2)$ is defined as follows

$$\begin{aligned} Q \doteq \frac{|\xi |(1+|\xi |)^{s}(|\xi _1||\xi _2|+|\xi _2||\xi -\xi _1-\xi _2|+|\xi -\xi _1-\xi _2||\xi _1|)(1+|\xi _1|)^{-s}(1+|\xi _2|)^{-s}(1+|\xi -\xi _1-\xi _2|)^{-s}}{(1+\xi ^2)(1+|\tau -\xi ^3|)^{1-b}(1+|\tau _1-\xi _1^3|)^{b'}(1+|\tau _2-\xi _2^3|)^{b'}(1+|\tau -\tau _1-\tau _2-(\xi -\xi _1-\xi _2)^3|)^{b'}}. \end{aligned}$$

Thanks to our symmetric writing (3.46), we can assume the following order of $|\xi _1|$, $|\xi _2|$ and $|\xi -\xi _1-\xi _2|$ (similar to assumption (3.13))

$$\begin{aligned} |\xi _1| \ge |\xi _2| \ge |\xi -\xi _1-\xi _2|. \end{aligned}$$

(3.48)

Thus by ordering relation (3.48), we get $|\xi _1||\xi _2|+|\xi _2||\xi -\xi _1-\xi _2|+|\xi -\xi _1-\xi _2||\xi _1|\le 3|\xi _1||\xi _2|$, which combined with $(1+\xi ^2)^{-1}(1+|\xi |)^{s}\simeq (1+|\xi |)^{s-2}$, reduces Q as follows

$$\begin{aligned} Q\lesssim & {} \widetilde{Q}\\\doteq & {} \frac{\chi _{|\xi _1|\ge |\xi _2|\ge |\xi -\xi _1-\xi _2|} |\xi |(1\!+\!|\xi |)^{s-2}|\xi _1\xi _2|(1\!+\!|\xi _1|)^{-s}(1\!+\!|\xi _2|)^{-s}(1\!+\!|\xi \!-\!\xi _1\!-\!\xi _2|)^{-s}}{(1\!+\!|\tau -\xi ^3|)^{1-b}(1\!+\!|\tau _1-\xi _1^3|)^{b'}(1\!+\!|\tau _2\!-\!\xi _2^3|)^{b'}(1\!+\!|\tau \!-\!\tau _1\!-\!\tau _2\!-\!(\xi \!-\!\xi _1\!-\!\xi _2)^3|)^{b'}}.\nonumber \end{aligned}$$

(3.49)

Furthermore, for $|\xi |\ge 10^{-3}|\xi _1|$, we have $|\xi _1\xi _2|\lesssim (1+|\xi |)^2$ or $(1+|\xi |)^{-2}|\xi _1\xi _2|\lesssim 1$, which gives

$$ \widetilde{Q} \lesssim \! \frac{\chi _{|\xi _1|\ge |\xi _2|\ge |\xi -\xi _1-\xi _2|}|\xi |(1\!+\!|\xi |)^{s}(1\!+\!|\xi _1|)^{-s}(1\!+\!|\xi _2|)^{-s}(1\!+\!|\xi \!-\!\xi _1\!-\!\xi _2|)^{-s}}{(1\!+\!|\tau \!-\!\xi ^3|)^{1-b}(1\!+\!|\tau _1\!-\!\xi _1^3|)^{b'}(1\!+\!|\tau _2\!-\!\xi _2^3|)^{b'}(1\!+\!|\tau \!-\!\tau _1\!-\!\tau _2\!-\!(\xi \!-\!\xi _1\!-\!\xi _2)^3|)^{b'}}, $$

where the right-hand side of the above inequality is bounded by the multiplier defined in (3.6). So, the $L^2$ inequality (3.47) is reduced to the $L^2$ inequality (3.5) for the mKdV trilinear estimate. Hence, we assume that $|\xi |< 10^{-3}|\xi _1|$ and the multiplier $\widetilde{Q}$ defined in (3.49) becomes

$$\begin{aligned} Q_1 \doteq \frac{\chi _{B}(\xi ,\xi _1,\xi _2) \cdot |\xi | (1\!+\!|\xi |)^{s-2}|\xi _1\xi _2|(1\!+\!|\xi _1|)^{-s} (1\!+\!|\xi _2|)^{-s}(1\!+\!|\xi \!-\!\xi _1\!-\!\xi _2|)^{-s}}{(1\!+\!|\tau \!-\!\xi ^3|)^{1-b} (1\!+\!|\tau _1\!-\!\xi _1^3|)^{b'}(1\!+\!|\tau _2\!-\!\xi _2^3|)^{b'} (1\!+\!|\tau \!-\!\tau _1\!-\!\tau _2\!-\!(\xi \!-\!\xi _1\!-\!\xi _2)^3|)^{b'}}, \end{aligned}$$

with the domain B given by

$$\begin{aligned} B \doteq \{(\xi ,\xi _1,\xi _2) \in \mathbb {R}^3: |\xi _1| \ge |\xi _2| \ge |\xi -\xi _1-\xi _2| ~\text { and }~|\xi |<10^{-3}|\xi _1|\}. \end{aligned}$$

(3.50)

Now, using duality, the left-hand side of $L^2$ form (3.47) (with $Q=Q_1$) is bounded as follows

$$\begin{aligned}{} & {} \left\| \int _{\mathbb {R}^4} Q_1 \cdot c_f(\xi _1,\tau _1) c_g(\xi _2,\tau _2) c_h(\xi -\xi _1-\xi _2,\tau -\tau _1-\tau _2) d\xi _2d\tau _2 d\xi _1d\tau _1 \right\| _{L^2_{\xi ,\tau }}\\\lesssim & {} \sup _{\Vert d\Vert _{L^2}=1}\!\! \int _{\mathbb {R}^2}\!\!\!\!\!\! d(\xi ,\tau )\!\!\left[ \!\!\int _{\mathbb {R}^4}\!\!\!\! Q_1c_f(\xi _1,\tau _1) c_g(\xi _2,\tau _2) c_h(\xi \!-\!\xi _1\!-\!\xi _2,\tau \!-\!\tau _1\!-\!\tau _2)d\xi _2d\tau _2d\xi _1d\tau _1\!\!\right] \!\!d\xi d\tau .\nonumber \end{aligned}$$

(3.51)

Furthermore, using Fubini’s Theorem in (3.51) to switch integrations $d\xi _1d\tau _1$ and $d\xi d\tau$, we get

$$\begin{aligned}{} & {} \left\| \int _{\mathbb {R}^4} Q_1 \cdot c_f(\xi _1,\tau _1) c_g(\xi _2,\tau _2)c_h(\xi -\xi _1-\xi _2,\tau -\tau _1-\tau _2)d\xi _2d\tau _2 d\xi _1d\tau _1\right\| _{L^2_{\xi ,\tau }}\\\lesssim & {} \sup _{\Vert d\Vert _{L^2}=1}\!\!\int _{\mathbb {R}^2} \!\!\!\!\!c_f(\xi _1,\tau _1)\!\!\left[ \!\!\int _{\mathbb {R}^4}\!\!\! Q_1 d(\xi ,\tau )c_g(\xi _2,\tau _2)c_h(\xi \!-\!\xi _1\!-\!\xi _2,\tau \!-\!\tau _1\!-\!\tau _2) d\xi _2d\tau _2 d\xi d\tau \!\right] \!\!d\xi _1 d\tau _1.\nonumber \end{aligned}$$

(3.52)

In addition, using Cauchy–Schwarz inequality for the integral $d\xi _1 d\tau _1$ in (3.52) we get

$$\begin{aligned}{} & {} \left\| \int _{\mathbb {R}^4} Q_1 \cdot c_f(\xi _1,\tau _1) c_g(\xi _2,\tau _2) c_h(\xi -\xi _1-\xi _2,\tau -\tau _1-\tau _2) d\xi _2d\tau _2 d\xi _1d\tau _1 \right\| _{L^2_{\xi ,\tau }}\\\lesssim & {} \sup _{\Vert d\Vert _{L^2}=1}\!\!\!\!\! \Vert c_f\Vert _{L^2}\!\!\!\left( \!\!\int _{\mathbb {R}^2}\!\!\left| \!\int _{\mathbb {R}^4}\!\!\!Q_1 d(\xi ,\tau ) c_g(\xi _2,\tau _2) c_h(\xi \!-\!\xi _1\!-\!\xi _2,\tau \!-\!\tau _1\!-\!\tau _2) d\xi _2d\tau _2 d\xi d\tau \right| ^2\!\!\! d\xi _1d\tau _1\!\!\!\right) ^{\!\!\frac{1}{2}}.\nonumber \end{aligned}$$

(3.53)

Moreover, for the $d\xi _2d\tau _2 d\xi d\tau$ integral in (3.53), applying the Cauchy–Schwarz inequality we get

$$\begin{aligned} \begin{aligned}&\left\| \int _{\mathbb {R}^4} Q_1 \cdot c_f(\xi _1,\tau _1) c_g(\xi _2,\tau _2) c_h(\xi -\xi _1-\xi _2,\tau -\tau _1-\tau _2) d\xi _2d\tau _2 d\xi _1d\tau _1\right\| _{L^2_{\xi ,\tau }}\\&\lesssim \sup _{\Vert d\Vert _{L^2}=1} \Vert c_f\Vert _{L^2} \left( \int _{\mathbb {R}^2}\left[ \int _{\mathbb {R}^4} Q_1^2(\xi ,\xi _1,\xi _2,\tau ,\tau _1,\tau _2) d\xi _2d\tau _2 d\xi d\tau \right] \right. \\&\quad \left. \times \left[ \int _{\mathbb {R}^4} d^2(\xi ,\tau ) c_g^2(\xi _2,\tau _2) c_h^2(\xi -\xi _1-\xi _2,\tau -\tau _1-\tau _2) d\xi _2d\tau _2 d\xi d\tau \right] d\xi _1d\tau _1\right) ^{1/2}. \end{aligned} \end{aligned}$$

(3.54)

In (3.54), taking the supremum of $\int _{\mathbb {R}^4} Q_1^2(\xi ,\xi _1,\xi _2,\tau ,\tau _1,\tau _2)d\xi _2d\tau _2 d\xi d\tau$ over $\xi _1$, $\tau _1$, we get

$$\begin{aligned}{} & {} \left\| \int _{\mathbb {R}^4} Q_1 \cdot c_f(\xi _1,\tau _1) c_g(\xi _2,\tau _2) c_h(\xi -\xi _1-\xi _2,\tau -\tau _1-\tau _2) d\xi _2d\tau _2 d\xi _1d\tau _1 \right\| _{L^2_{\xi ,\tau }}\nonumber \\\lesssim & {} \sup _{\Vert d\Vert _{L^2}=1} \Vert c_f\Vert _{L^2} \left\| \int _{\mathbb {R}^4} Q_1^2(\xi ,\xi _1,\xi _2,\tau ,\tau _1,\tau _2) d\xi _2d\tau _2 d\xi d\tau \right\| _{L^\infty _{\xi _1,\tau _1}}^{\frac{1}{2}}\\{} & {} \times \left( \int _{\mathbb {R}^6} d^2(\xi ,\tau ) c_g^2(\xi _2,\tau _2) c_h^2(\xi -\xi _1-\xi _2,\tau -\tau _1-\tau _2) d\xi _2d\tau _2 d\xi d\tau d\xi _1d\tau _1 \right) ^{1/2}.\nonumber \end{aligned}$$

(3.55)

Next, using the following estimate (which similar to (3.11))

$$\begin{aligned}{} & {} \int _{\mathbb {R}^6} d^2(\xi ,\tau ) c_g^2(\xi _2,\tau _2) c_h^2(\xi -\xi _1-\xi _2,\tau -\tau _1-\tau _2) d\xi _2d\tau _2 d\xi d\tau d\xi _1d\tau _1\nonumber \\{} & {} \lesssim \Vert d\Vert _{L^2}^2 \Vert c_g\Vert _{L^2}^2 \Vert c_h\Vert _{L^2}^2, \end{aligned}$$

from estimate (3.55) we arrive at

$$\begin{aligned}{} & {} \left\| \int _{\mathbb {R}^4} Q_1 \cdot c_f(\xi _1,\tau _1) c_g(\xi _2,\tau _2) c_h(\xi -\xi _1-\xi _2,\tau -\tau _1-\tau _2) d\xi _2d\tau _2 d\xi _1d\tau _1\right\| _{L^2_{\xi ,\tau }}\nonumber \\\lesssim & {} \sup _{\Vert d\Vert _{L^2}=1} \Vert c_f\Vert _{L^2} \left\| \int _{\mathbb {R}^4}Q_1^2(\xi ,\xi _1,\xi _2,\tau ,\tau _1,\tau _2)d\xi _2d\tau _2 d\xi d\tau \right\| _{L^\infty _{\xi _1,\tau _1}}^{\frac{1}{2}}\Vert d\Vert _{L^2}\Vert c_g\Vert _{L^2}\Vert c_h\Vert _{L^2}\nonumber \\\lesssim & {} \Vert {\varTheta }_2\Vert _{L_{\xi _1,\tau _1}^\infty }^{\frac{1}{2}}\Vert c_f\Vert _{L^2_{\xi ,\tau }}\Vert c_g\Vert _{L^2_{\xi ,\tau }}\Vert c_h\Vert _{L^2_{\xi ,\tau }}, \end{aligned}$$

(3.56)

where ${\varTheta }_2$ is as in the following result, which provides its estimate.

Lemma 3.4

If $s> \frac{1}{2}$ and $\frac{1}{2}<b'\le b\le \min \{\frac{s}{3}+\frac{1}{3},1\}$, then for all $\xi _1$ and $\tau _1$ we have

$$\begin{aligned}{} & {} {\varTheta }_2(\xi _1,\tau _1)\doteq \frac{(1+|\xi _1|)^{-2s}}{(1+|\tau _1-\xi _1^3|)^{2b'}}\nonumber \\{} & {} \times \!\! \int _{\mathbb {R}^4}\!\!\! \frac{\chi _{B}(\xi ,\xi _1,\xi _2) \cdot |\xi |^2(1\!+\!|\xi |)^{2s-4}|\xi _1\xi _2|^2 (1\!+\!|\xi _2|)^{-2s} (1\!+\!|\xi \!-\!\xi _1\!-\!\xi _2|)^{-2s}d\tau _2d\xi _2d\tau d\xi }{(1\!+\!|\tau -\xi ^3|)^{2-2b}(1+|\tau _2-\xi _2^3|)^{2b'} (1+|\tau -\tau _1-\tau _2-(\xi -\xi _1-\xi _2)^3|)^{2b'}}\nonumber \\\lesssim & {} 1. \end{aligned}$$

(3.57)

Proof

For the $d\tau _2$-integral in (3.57), applying estimate (3.1) with $a=\tau -\tau _1-(\xi -\xi _1-\xi _2)^3$, $c=\xi _2^3$, $\ell =b'$, we get

$$\begin{aligned} {\varTheta }_2\lesssim & {} \frac{(1+|\xi _1|)^{-2s}}{(1+|\tau _1-\xi _1^3|)^{2b'}}\\{} & {} \times \int _{\mathbb {R}^3} \frac{\chi _{B}\cdot |\xi |^2(1+|\xi |)^{2s-4}|\xi _1\xi _2|^2 (1+|\xi _2|)^{-2s}(1+|\xi -\xi _1-\xi _2|)^{-2s}d\tau d\xi _2 d\xi }{(1+|\tau -\xi ^3|)^{2(1-b)}(1+|\tau -\tau _1-(\xi -\xi _1-\xi _2)^3-\xi _2^3|)^{2b'}}.\nonumber \end{aligned}$$

(3.58)

Furthermore, for the $d\tau$-integral in (3.58), applying estimate (3.2) with $c=\tau _1+(\xi -\xi _1-\xi _2)^3+\xi _2^3$, $a=\xi ^3$, $\ell =b$, $\ell '=b'$ we get

$$\begin{aligned} {\varTheta }_2\lesssim & {} \frac{(1+|\xi _1|)^{-2s}}{(1+|\tau _1-\xi _1^3|)^{2b'}}\\{} & {} \times \int _{\mathbb {R}^2} \frac{\chi _{B} |\xi |^2 (1+|\xi |)^{2s-4} |\xi _1\xi _2|^2 (1+|\xi _2|)^{-2s}(1+|\xi -\xi _1-\xi _2|)^{-2s}d\xi _2 d\xi }{(1+|\tau _1-\xi _1^3+d_3(\xi ,\xi _1,\xi _2)|)^{2(1-b)}},\nonumber \end{aligned}$$

(3.59)

where $d_3(\xi ,\xi _1,\xi _2)$ is the Bourgain quantity defined by (3.8). Next, we consider the following cases.

• Case 1: $|\xi _1|\le 100$.

• Case 2: $|\xi _1|> 100$.

Proof in Case 1. Since $|\xi |<10^{-3}|\xi _1|$ and $|\xi _1|\le 100$, by the ordering relation (3.48), all of $|\xi |$, $|\xi _1|$ and $|\xi _2|$ are bounded. Furthermore, since $b'\ge 0$ and $(1-b)\ge 0$, we see that the multiplier and the integrand in ${\varTheta }_2$ are bounded. Therefore, ${\varTheta }_2(\xi _1,\tau _1)\lesssim 1$, since the integration is over a bounded set.

Proof in Case 2. Here we have

$$\begin{aligned} |\xi _2| \simeq |\xi _1|. \end{aligned}$$

(3.60)

By the ordering relation $|\xi _1| \ge |\xi _2| \ge |\xi -\xi _1-\xi _2|$, it suffices to show $|\xi _2|> \frac{1}{4} |\xi _1|$. In fact, if $|\xi _2|\le \frac{1}{4}|\xi _1|$, then we would get $|\xi |= |\xi _1+\xi _2+(\xi -\xi _1-\xi _2)| \ge |\xi _1|-|\xi -\xi _1-\xi _2|-|\xi _2| \ge |\xi _1|-\frac{1}{4}|\xi _1|-\frac{1}{4}|\xi _1| \ge \frac{1}{2}|\xi _1|$, which is a contradiction to $|\xi |<10^{-3}|\xi _1|$, a condition of $(\xi ,\xi _1,\xi _2)\in B$. Using (3.60) we get $|\xi _1\xi _2|^2\simeq |\xi _1|^4$, and $(1+|\xi _2|)^{-2s}\simeq |\xi _1|^{-2s}$ since $|\xi _1|>100$. Also, we have $(1+|\xi _1|)^{-2s}\simeq |\xi _1|^{-2s}$. Combining this with (3.59) we get

$$\begin{aligned} \begin{aligned} {\varTheta }_2(\xi _1,\tau _1)&\lesssim \frac{|\xi _1|^{4-4s}}{(1+|\tau _1-\xi _1^3|)^{2b'}} \\&\int _{\mathbb {R}^2} \frac{\chi _{B}(\xi ,\xi _1,\xi _2)\cdot (1+|\xi |)^{2s-2} (1+|\xi -\xi _1-\xi _2|)^{-2s} }{(1+|\tau _1-\xi _1^3+d_3(\xi ,\xi _1,\xi _2)|)^{2(1-b)}} d\xi _2 d\xi . \end{aligned} \end{aligned}$$

(3.61)

Since $\frac{1}{2}<b'\le b<1$ we have $0<2(1-b)<1<2b'$, so we can move $(1+|\tau _1-\xi _1^3|)^{2b'}$ inside the integral and replace $2b'$ with $2(1-b)$. Since $(1+|\tau _1-\xi _1^3|)(1+|\tau _1-\xi _1^3+$ $d_3(\xi ,\xi _1,\xi _2)|) \ge |\tau _1-\xi _1^3|+|\tau _1-\xi _1^3+d_3(\xi ,\xi _1,\xi _2)|$ and also $|\tau _1-\xi _1^3|+ |\tau _1-\xi _1^3+$ $d_3(\xi ,\xi _1,\xi _2)| \ge |\tau _1-\xi _1^3+d_3(\xi ,\xi _1,\xi _2)- (\tau _1-\xi _1^3)|= |d_3(\xi ,\xi _1,\xi _2)|$, we have

$$\begin{aligned} (1+|\tau _1-\xi _1^3|)^{2b'} (1+ |\tau _1-\xi _1^3+d_3(\xi ,\xi _1,\xi _2)|)^{2(1-b)} \gtrsim |d_3(\xi ,\xi _1,\xi _2)|^{2(1-b)}, \end{aligned}$$

which combined with (3.61) gives

$$\begin{aligned} {\varTheta }_2(\xi _1,\tau _1) \lesssim |\xi _1|^{4-4s} \int _{\mathbb {R}^2}\frac{ \chi _{B}(\xi ,\xi _1,\xi _2) \cdot (1+|\xi |)^{2s-2} (1+|\xi -\xi _1-\xi _2|)^{-2s} }{|d_3(\xi ,\xi _1,\xi _2)|^{2(1-b)}}d\xi _2 d\xi . \end{aligned}$$

Combining this with property (3.8), i.e. $d_3 = -3(\xi -\xi _1)(\xi -\xi _2)(\xi _1+\xi _2)$, we get

$$\begin{aligned} {\varTheta }_2(\xi _1,\tau _1) \lesssim |\xi _1|^{4-4s} \int _{\mathbb {R}^2} \frac{\chi _{B}(\xi ,\xi _1,\xi _2)\cdot (1+|\xi |)^{2s-2} (1+|\xi -\xi _1-\xi _2|)^{-2s}}{|(\xi -\xi _1)(\xi -\xi _2)(\xi _1+\xi _2)|^{2(1-b)}}d\xi _2 d\xi . \end{aligned}$$

(3.62)

Next, using $(1+|\xi -\xi _1-\xi _2|)^{-2s}\lesssim 1$ and the inequalities $|\xi -\xi _1|\gtrsim |\xi _1|$ and $|\xi -\xi _2| \gtrsim |\xi _1|$, which follows from $|\xi -\xi _1| \ge |\xi _1|-|\xi | \ge |\xi _1|-10^{-3}|\xi _1| \gtrsim |\xi _1|$ and $|\xi -\xi _2| \ge |\xi _2|-|\xi | \ge \frac{1}{4}|\xi _1|-10^{-3}|\xi _1| \gtrsim |\xi _1|$, from (3.62) we obtain

$$\begin{aligned} \begin{aligned} {\varTheta }_2(\xi _1,\tau _1)&\lesssim |\xi _1|^{4b-4s} \int _{\mathbb {R}^2} \frac{\chi _{B} (\xi ,\xi _1,\xi _2) \cdot (1+|\xi |)^{2s-2}}{|\xi _1+\xi _2|^{2(1-b)}} d\xi _2 d\xi \\&= |\xi _1|^{4b-4s} \int _{|\xi |\le 10^{-3}|\xi _1|} \frac{d\xi }{(1+|\xi |)^{2-2s}}\int _{|\xi _2|\le |\xi _1|} \frac{d\xi _2}{|\xi _1+\xi _2|^{2(1-b)}}. \end{aligned} \end{aligned}$$

(3.63)

Since $s>\frac{1}{2}$, we have $2-2s<1$, which implies the following bound for the first integral

$$\begin{aligned} \int _{|\xi |\le 10^{-3}|\xi _1|} \frac{d\xi }{(1+|\xi |)^{2-2s}} \simeq \left[ (1+\xi )^{2s-1}\right] \big |_0^{10^{-3} |\xi _1|} \lesssim |\xi _1|^{2s-1}. \end{aligned}$$

(3.64)

Concerning the second integral in (3.63), we make the change of variables $\mu =\mu (\xi _2)=\xi _1+\xi _2$, and using the inequalities $|\mu |\le |\xi _1|+|\xi _2|\le 2|\xi _1|$, for $2(1-b)<1$ or $b>\frac{1}{2}$, we have

$$\begin{aligned} \int _{|\xi _2|\le |\xi _1|} \frac{d\xi _2}{|\xi _1+\xi _2|^{2(1-b)}} = \int _{|\mu |\le 2 |\xi _1|} \frac{d\mu }{|\mu |^{2(1-b)}} \simeq (\mu ^{2b-1})\big |_0^{2|\xi _1|} \simeq |\xi _1|^{2b-1}. \end{aligned}$$

(3.65)

Combining estimates (3.64), (3.65) with (3.63), we obtain

$$\begin{aligned} {\varTheta }_2(\xi _1,\tau _1) \lesssim |\xi _1|^{4b-4s} |\xi _1|^{2s-1} |\xi _1|^{2b-1} = |\xi _1|^{6b-2s-2}. \end{aligned}$$

Since $|\xi _1|>100$, the above quantity is bounded if $6b-2s-2\le 0$ or $b\le \frac{1}{3}s+\frac{1}{3}$. For $b>\frac{1}{2}$, it suffices to have $\frac{1}{3}s+\frac{1}{3}>\frac{1}{2}$ or $\frac{1}{3}s>\frac{1}{6}$, which implies that $s>1/2$. This completes the proof of Lemma 3.4. $\square$

3.3 Proof of the Second Nonlocal Trilinear Estimates (2.18)

Using notation (3.4), we see that to prove the estimate (2.18), i.e.

$$ \Vert (1+\partial _x^2)^{-1}[(\partial _xf)(\partial _xg)(\partial _xh)]\Vert _{X^{s,b-1}} \le c_{s,b}\Vert f\Vert _{X^{s,b'}}\Vert g\Vert _{X^{s,b'}}\Vert h\Vert _{X^{s,b'}}, $$

it suffices to prove the following $L^2$ estimate

$$\begin{aligned} \begin{aligned}&\left\| \int _{\mathbb {R}^4} Q \cdot c_f(\xi _1,\tau _1) c_g(\xi _2,\tau _2) c_h(\xi -\xi _1-\xi _2,\tau -\tau _1-\tau _2) d\xi _2d\tau _2 d\xi _1d\tau _1\right\| _{L^2_{\xi ,\tau }}\\&\lesssim \Vert c_f\Vert _{L^2_{\xi ,\tau }}\Vert c_g\Vert _{L^2_{\xi ,\tau }}\Vert c_h\Vert _{L^2_{\xi ,\tau }}, \end{aligned} \end{aligned}$$

(3.66)

where the multiplier $Q = Q(\xi ,\tau ,\xi _1,\tau _1,\xi _2,\tau _2)$ is defined by

$$\begin{aligned} Q \doteq \frac{(1\!+\!\xi ^2)^{-1}(1\!+\!|\xi |)^{s} |\xi _1||\xi _2||\xi \!-\!\xi _1\!-\!\xi _2| (1\!+\!|\xi _1|)^{-s} (1\!+\!|\xi _2|)^{-s}(1\!+\!|\xi \!-\!\xi _1\!-\!\xi _2|)^{-s}}{(1\!+\!|\tau \!-\!\xi ^3|)^{1-b} (1\!+\!|\tau _1\!-\!\xi _1^3|)^{b'} (1\!+\!|\tau _2\!-\!\xi _2^3|)^{b'} (1\!+\!|\tau \!-\!\tau _1\!-\!\tau _2\!-\!(\xi \!-\!\xi _1\!-\!\xi _2)^3|)^{b'}}. \end{aligned}$$

By symmetry in convolution writing of $\widehat{\partial _x f} *\widehat{\partial _xg} *\widehat{\partial _x h}$, we can assume the following order of $|\xi _1|$, $|\xi _2|$ and $|\xi -\xi _1-\xi _2|$

$$\begin{aligned} |\xi _1| \ge |\xi _2| \ge |\xi -\xi _1-\xi _2|. \end{aligned}$$

(3.67)

Therefore, the multiplier Q is reduced to $\chi _{|\xi _1|\ge |\xi _2|\ge |\xi -\xi _1-\xi _2|}Q$. Also, using $(1+\xi ^2)^{-1}(1+|\xi |)^{s}\lesssim (1+|\xi |)^{s-2}$, Q becomes

$$\begin{aligned} Q \lesssim \widetilde{Q}\doteq \frac{\chi _{|\xi _1|\ge |\xi _2|\ge |\xi \!-\!\xi _1\!-\!\xi _2|} (1\!+\!|\xi |)^{s-2} |\xi _1\xi _2(\xi \!-\!\xi _1\!-\!\xi _2)| (1\!+\!|\xi _1|)^{-s} (1\!+\!|\xi _2|)^{-s} (1\!+\!|\xi \!-\!\xi _1\!-\!\xi _2|)^{-s}}{(1\!+\!|\tau \!-\!\xi ^3|)^{1-b}(1\!+\!|\tau _1\!-\!\xi _1^3|)^{b'} (1\!+\!|\tau _2\!-\!\xi _2^3|)^{b'} (1\!+\!|\tau \!-\!\tau _1\!-\!\tau _2\!-\!(\xi \!-\!\xi _1\!-\!\xi _2)^3|)^{b'}}. \end{aligned}$$

Furthermore, for $|\xi |\ge 10^{-3}|\xi _1|$, we have $|\xi _1\xi _2(\xi -\xi _1-\xi _2)|\lesssim |\xi |(1+|\xi |)^2$ or $(1+|\xi |)^{-2} |\xi _1\xi _2(\xi -\xi _1-\xi _2)|\lesssim |\xi |$, which implies that

$$\begin{aligned} \widetilde{Q} \lesssim \frac{\chi _{|\xi _1|\ge |\xi _2|\ge |\xi \!-\!\xi _1\!-\!\xi _2|} |\xi |(1\!+\!|\xi |)^{s}(1\!+\!|\xi _1|)^{-s}(1\!+\!|\xi _2|)^{-s}(1\!+\!|\xi -\xi _1\!-\!\xi _2|)^{-s}}{(1\!+\!|\tau \!-\!\xi ^3|)^{1-b}(1\!+\!|\tau _1\!-\!\xi _1^3|)^{b'} (1\!+\!|\tau _2\!-\!\xi _2^3|)^{b'} (1\!+\!|\tau \!-\!\tau _1\!-\!\tau _2\!-\!(\xi \!-\!\xi _1\!-\!\xi _2)^3|)^{b'}}, \end{aligned}$$

where the right-hand side of the above inequality is bounded by the multiplier defined in (3.6). So, the $L^2$ inequality (3.66) is reduced to the $L^2$ inequality (3.5) for the local trilinear estimate. Hence, we assume that $|\xi |< 10^{-3}|\xi _1|$ and the multiplier $\widetilde{Q}$ defined in (3.49) becomes

$$\begin{aligned} Q_2 \doteq \frac{\chi _{B}(\xi ,\xi _1,\xi _2) \cdot (1\!+\!|\xi |)^{s-2} |\xi _1\xi _2(\xi \!-\!\xi _1\!-\!\xi _2)| (1\!+\!|\xi _1|)^{-s} (1\!+\!|\xi _2|)^{-s} (1\!+\!|\xi \!-\!\xi _1\!-\!\xi _2|)^{-s}}{(1\!+\!|\tau \!-\!\xi ^3|)^{1-b} (1\!+\!|\tau _1\!-\!\xi _1^3|)^{b'} (1\!+\!|\tau _2\!-\!\xi _2^3|)^{b'} (1\!+\!|\tau \!-\!\tau _1\!-\!\tau _2\!-\!(\xi \!-\!\xi _1\!-\!\xi _2)^3|)^{b'}}, \end{aligned}$$

where B is the domain defined by (3.50) incorporating the order relation and the condition $|\xi |<10^{-3}|\xi _1|$. Like in the proof for the first nonlocal trilinear estimate (see (3.56)), using duality we bound the left-hand side in the $L^2$ formulation (3.66) (with $Q=Q_2$) as follows

$$\begin{aligned}{} & {} \left\| \int _{\mathbb {R}^4} Q_2 \cdot c_f(\xi _1,\tau _1) c_g(\xi _2,\tau _2)c_h(\xi -\xi _1-\xi _2,\tau -\tau _1-\tau _2) d\xi _2d\tau _2 d\xi _1d\tau _1 \right\| _{L^2}\\{} & {} \lesssim \Vert {\varTheta }_3\Vert _{L_{\xi _1,\tau _1}^\infty }^{\frac{1}{2}}\Vert c_f\Vert _{L^2}\Vert c_g\Vert _{L^2}\Vert c_h\Vert _{L^2}, \end{aligned}$$

where ${\varTheta }_3$ is as in the following result, which provides its $L^\infty$ estimate.

Lemma 3.5

If $s> \frac{2}{3}$ and $\frac{1}{2}<b'\le b< \min \{s-\frac{1}{6}, \frac{s}{3}+\frac{1}{3}, \frac{5}{6}\}$, then for all $\xi _1$ and $\tau _1$ we have

$$\begin{aligned}{} & {} {\varTheta }_3(\xi _1,\tau _1) \doteq \frac{(1+|\xi _1|)^{-2s}}{(1+|\tau _1-\xi _1^3|)^{2b'}}\nonumber \\{} & {} \times \!\! \int _{\mathbb {R}^4}\!\!\!\! \frac{\chi _{B}(\xi ,\xi _1,\xi _2)(1\!+\!|\xi |)^{2s-4} |\xi _1\xi _2(\xi \!-\!\xi _1\!-\!\xi _2)|^2 (1\!+\!|\xi _2|)^{-2s}(1\!+\!|\xi \!-\!\xi _1\!-\!\xi _2|)^{-2s} d\tau _2d\xi _2d\tau d\xi }{(1+|\tau -\xi ^3|)^{2-2b} (1+|\tau _2-\xi _2^3|)^{2b'}(1+|\tau -\tau _1-\tau _2-(\xi -\xi _1-\xi _2)^3|)^{2b'}}\nonumber \\\lesssim & {} 1. \end{aligned}$$

(3.68)

Proof

For the $d\tau _2$-integral in (3.68), applying estimate (3.1) with $a=\tau -\tau _1-(\xi -\xi _1-\xi _2)^3$, $c=\xi _2^3$, $\ell =b'$, we get

$$\begin{aligned} {\varTheta }_3\lesssim & {} \frac{(1+|\xi _1|)^{-2s}}{(1+|\tau _1-\xi _1^3|)^{2b'}}\\{} & {} \times \int _{\mathbb {R}^3}\frac{\chi _{B}(\xi ,\xi _1,\xi _2)(1+|\xi |)^{2s-4}|\xi _1\xi _2(\xi -\xi _1-\xi _2)|^2(1+|\xi _2|)^{-2s}(1+|\xi -\xi _1-\xi _2|)^{-2s}\,d\tau d\xi _2 d\xi }{(1+|\tau -\xi ^3|)^{2(1-b)}(1+|\tau -\tau _1-(\xi -\xi _1-\xi _2)^3-\xi _2^3|)^{2b'}}. \end{aligned}$$

Furthermore, for the $d\tau$-integral in the above estimate, applying estimate (3.2) with $c=\tau _1+(\xi -\xi _1-\xi _2)^3+\xi _2^3$, $a=\xi ^3$, $\ell =b$, $\ell '=b'$, we get

$$\begin{aligned} {\varTheta }_3\lesssim & {} \frac{(1+|\xi _1|)^{-2s}}{(1+|\tau _1-\xi _1^3|)^{2b'}}\nonumber \\{} & {} \times \int _{\mathbb {R}^2}\frac{\chi _{B}(1+|\xi |)^{2s-4}|\xi _1\xi _2(\xi -\xi _1-\xi _2)|^2(1+|\xi _2|)^{-2s}(1+|\xi -\xi _1-\xi _2|)^{-2s}d\xi _2 d\xi }{(1+|\tau _1+(\xi -\xi _1-\xi _2)^3+\xi _2^3-\xi ^3|)^{2(1-b)}}\nonumber \\= & {} \frac{(1+|\xi _1|)^{-2s}}{(1+|\tau _1-\xi _1^3|)^{2b'}}\\{} & {} \times \int _{\mathbb {R}^2}\frac{\chi _{B} (1+|\xi |)^{2s-4} |\xi _1\xi _2(\xi -\xi _1-\xi _2)|^2 (1+|\xi _2|)^{-2s} (1+|\xi -\xi _1-\xi _2|)^{-2s}}{(1+|\tau _1-\xi _1^3+d_3(\xi ,\xi _1,\xi _2)|)^{2(1-b)}}d\xi _2 d\xi ,\nonumber \end{aligned}$$

(3.69)

where $d_3(\xi ,\xi _1,\xi _2)$ is the Bourgain quantity defined by (3.7). Next, we consider the following cases.

• Case 1: $|\xi _1|\le 100$.

• Case 2: $|\xi _1|> 100$.

Proof in Case 1. Since $|\xi |<10^{-3}|\xi _1|$ and $|\xi _1|\le 100$, by the ordering relation (3.67), all of $|\xi |$, $|\xi _1|$ and $|\xi _2|$ are bounded. Furthermore, since $b'\ge 0$ and $(1-b)\ge 0$, we see that ${\varTheta }_3(\xi _1,\tau _1)\lesssim 1$.

Proof in Case 2. Here we have

$$\begin{aligned} |\xi _2| \simeq |\xi _1|. \end{aligned}$$

(3.70)

By the ordering relation $|\xi _1| \ge |\xi _2| \ge |\xi -\xi _1-\xi _2|$, it suffices to show $|\xi _2|> \frac{1}{4} |\xi _1|$. In fact, if $|\xi _2|\le \frac{1}{4}|\xi _1|$, then we would get $|\xi | = |\xi _1+\xi _2+(\xi -\xi _1-\xi _2)| \ge |\xi _1|-|\xi -\xi _1-\xi _2|-|\xi _2| \ge |\xi _1|-\frac{1}{4}|\xi _1|-\frac{1}{4}|\xi _1| \ge \frac{1}{2}|\xi _1|$, which is a contradiction to $|\xi |<10^{-3}|\xi _1|$, a condition of $(\xi ,\xi _1,\xi _2)\in B$. Using (3.70) we get $|\xi _1\xi _2|^2\simeq |\xi _1|^4$, and $(1+|\xi _2|)^{-2s}\simeq |\xi _1|^{-2s}$ since $|\xi _1|>100$. Also, we have $(1+|\xi _1|)^{-2s}\simeq |\xi _1|^{-2s}$. Combining this with $|\xi -\xi _1-\xi _2|^2\lesssim (1+|\xi -\xi _1-\xi _2|)^{2}$, from (3.69) we get

$$\begin{aligned} \begin{aligned} {\varTheta }_3(\xi _1,\tau _1)&\lesssim \frac{|\xi _1|^{4-4s}}{(1+|\tau _1-\xi _1^3|)^{2b'}}\\&\times \int _{\mathbb {R}^2} \frac{\chi _{B}(\xi ,\xi _1,\xi _2)\cdot (1+|\xi |)^{2s-4} (1+|\xi -\xi _1-\xi _2|)^{2-2s}}{(1+|\tau _1-\xi _1^3 +d_3(\xi ,\xi _1,\xi _2) |)^{2(1-b)}}d\xi _2 d\xi . \end{aligned} \end{aligned}$$

(3.71)

Since $\frac{1}{2}<b'\le b<1$ we have $0<2(1-b)<1<2b'$, so we can move $(1+|\tau _1-\xi _1^3|)^{2b'}$ inside the integral and replace $2b'$ with $2(1-b)$. Since $(1+|\tau _1-\xi _1^3|) (1+|\tau _1-\xi _1^3+$ $d_3(\xi ,\xi _1,\xi _2)|) \ge |\tau _1-\xi _1^3|+|\tau _1-\xi _1^3+d_3(\xi ,\xi _1,\xi _2)|$ and also $|\tau _1-\xi _1^3| + |\tau _1-\xi _1^3+$ $d_3(\xi ,\xi _1,\xi _2)| \ge |\tau _1-\xi _1^3+d_3(\xi ,\xi _1,\xi _2) - (\tau _1-\xi _1^3)| = |d_3(\xi ,\xi _1,\xi _2)|$, we have

$$\begin{aligned} (1+|\tau _1-\xi _1^3|)^{2b'} (1+ |\tau _1-\xi _1^3+d_3(\xi ,\xi _1,\xi _2)| )^{2(1-b)} \gtrsim |d_3(\xi ,\xi _1,\xi _2)|^{2(1-b)}, \end{aligned}$$

which combined with (3.71) gives

$$\begin{aligned} {\varTheta }_3(\xi _1,\tau _1) \lesssim |\xi _1|^{4-4s} \int _{\mathbb {R}^2} \frac{\chi _{B}(\xi ,\xi _1,\xi _2) \cdot (1+|\xi |)^{2s-4} (1+|\xi -\xi _1-\xi _2|)^{2-2s}}{|d_3(\xi ,\xi _1,\xi _2)|^{2(1-b)}}d\xi _2d\xi . \end{aligned}$$

Then, using the factorization $d_3 = -3(\xi -\xi _1)(\xi -\xi _2)(\xi _1+\xi _2)$, we get

$$\begin{aligned} {\varTheta }_3(\xi _1,\tau _1) \lesssim |\xi _1|^{4-4s} \int _{\mathbb {R}^2}\frac{\chi _{B}(\xi ,\xi _1,\xi _2)\cdot (1+|\xi |)^{2s-4} (1+|\xi -\xi _1-\xi _2|)^{2-2s}}{|(\xi -\xi _1) (\xi -\xi _2) (\xi _1+\xi _2)|^{2(1-b)}} d\xi _2 d\xi . \end{aligned}$$

(3.72)

Next, using the inequalities

$$ |\xi -\xi _1| \gtrsim |\xi _1| \quad \text { and }\quad |\xi -\xi _2|\gtrsim |\xi _1|, $$

which follows from $|\xi -\xi _1| \ge |\xi _1|-|\xi | \ge |\xi _1|-10^{-3}|\xi _1| \gtrsim |\xi _1|$ and $|\xi -\xi _2| \ge |\xi _2|-|\xi | \ge \frac{1}{4}|\xi _1|-10^{-3}|\xi _1| \gtrsim |\xi _1|$, from (3.72), we obtain

$$\begin{aligned} {\varTheta }_3(\xi _1,\tau _1) \lesssim |\xi _1|^{4b-4s}\!\!\!\int _{\mathbb {R}^2}\!\!\! \frac{\chi _{B}(\xi ,\xi _1,\xi _2) \cdot (1+|\xi |)^{2s-4} (1+|\xi -\xi _1-\xi _2|)^{2-2s}}{|\xi _1+\xi _2|^{2(1-b)}} d\xi _2 d\xi . \end{aligned}$$

(3.73)

Now, we consider the following two subcases arising from the sign of $(2-2s)$.

• Subcase 2.1: $s\le 1$.

• Subcase 2.2: $s>1$.

Proof in Subcase 2.1. Then $2-2s\ge 0$, which combined with $|\xi -\xi _1-\xi _2|\le |\xi _1|$ and $|\xi _1|>100$ gives $(1+|\xi -\xi _1-\xi _2|)^{2-2s}\lesssim |\xi _1|^{2-2s}$. From the last inequality and estimate (3.73) we get

$$\begin{aligned} \begin{aligned} {\varTheta }_3(\xi _1,\tau _1)&\lesssim |\xi _1|^{4b-6s+2} \int _{\mathbb {R}^2} \frac{\chi _{B}(\xi ,\xi _1,\xi _2) \cdot (1+|\xi |)^{2s-4}}{|\xi _1+\xi _2|^{2(1-b)}} d\xi _2 d\xi \\&= |\xi _1|^{4b-6s+2} \int _{|\xi |\le 10^{-3} |\xi _1|} \frac{d\xi }{(1+|\xi |)^{4-2s}} \cdot \int _{|\xi _2|\le |\xi _1|}\frac{d\xi _2}{|\xi _1+\xi _2|^{2(1-b)}}. \end{aligned} \end{aligned}$$

(3.74)

Since $4-2s\ge 2$ the first integral is bounded. For the second integral, making the change of variables $\mu =\mu (\xi _2)=\xi _1+\xi _2$ and using the inequalities $|\mu |\le |\xi _1|+|\xi _2|\le 2 |\xi _1|$, for $2(1-b)<1$ or $b>\frac{1}{2}$, we have $\int _{|\xi _2|\le |\xi _1|}\frac{d\xi _2}{|\xi _1+\xi _2|^{2(1-b)}}=\int _{|\mu |\le 2 |\xi _1|}\frac{d\mu }{|\mu |^{2(1-b)}}\simeq (\mu ^{2b-1})\big |_0^{2|\xi _1|}\simeq |\xi _1|^{2b-1}$. Using the above computations, from (3.74) we get

$$\begin{aligned} {\varTheta }_3(\xi _1,\tau _1) \lesssim |\xi _1|^{4b-6s+2} |\xi _1|^{2b-1} = |\xi _1|^{6b-6s+1}. \end{aligned}$$

Since $|\xi _1|>100$, the above quantity is bounded if $6b-6s+1\le 0$ or $b\le s-\frac{1}{6}$. For $b>\frac{1}{2}$, it suffices to have $s-\frac{1}{6}>\frac{1}{2}$ or $s>2/3$. This completes the proof in Subcase 2.1.

Proof in Subcase 2.2. Then $2-2s<0$, which implies that $(1+|\xi -\xi _1-\xi _2|)^{2-2s}\lesssim 1$. Therefore,

$$\begin{aligned} {\varTheta }_3(\xi _1,\tau _1) \lesssim |\xi _1|^{4b-4s} \int _{|\xi |\le 10^{-3} |\xi _1|} \frac{d\xi }{(1+|\xi |)^{4-2s}} \cdot \int _{|\xi _2|\le |\xi _1|} \frac{d\xi _2}{|\xi _1+\xi _2|^{2(1-b)}}. \end{aligned}$$

(3.75)

For the first integral, we have

$$\begin{aligned} \int _{|\xi |\le 10^{-3} |\xi _1|} \frac{d\xi }{(1+|\xi |)^{4-2s}} \lesssim \left\{ \begin{array}{ll} 1,&{}\quad s<\frac{3}{2},\\ \ln |\xi _1|,&{} \quad s=\frac{3}{2},\\ |\xi _1|^{2s-3},&{} \quad s>\frac{3}{2}. \end{array}\right. \end{aligned}$$

For the second integral in (3.75), making the change of variables $\mu =\mu (\xi _2)=\xi _1+\xi _2$ and using the inequalities $|\mu |\le |\xi _1|+|\xi _2|\le 2 |\xi _1|$, for $2(1-b)<1$ or $b>\frac{1}{2}$, we get $\int _{|\xi _2|\le |\xi _1|} \frac{d\xi _2}{|\xi _1+\xi _2|^{2(1-b)}} = \int _{|\mu |\le 2 |\xi _1|} \frac{d\mu }{|\mu |^{2(1-b)}} \simeq (\mu ^{2b-1})\big |_0^{2|\xi _1|} \simeq |\xi _1|^{2b-1}$. Finally, using the above computations, from (3.75) we get

$$\begin{aligned} {\varTheta }_3(\xi _1,\tau _1) \lesssim \left\{ \begin{array}{ll} |\xi _1|^{6b-4s-1},&{}\quad s<\frac{3}{2},\\ |\xi _1|^{6b-7}\ln |\xi _1|,&{}\quad s=\frac{3}{2},\\ |\xi _1|^{6b-2s-4},&{}\quad s>\frac{3}{2}, \end{array}\right. \end{aligned}$$

which is bounded if $b\le \frac{5}{6}$, since $s>1$. This completes the proof for Lemma 3.5, and also the proof of our last trilinear estimate. $\square$